Dependence Logic: A New Approach to Independence Friendly Logic (London Mathematical Society Student Texts)

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Dependence Logic Dependence is a common phenomenon, wherever one looks: ecological systems, astronomy, human history, stock markets – but what is the logic of dependence? This book is the first to carry out a systematic logical study of this important concept, giving on the way a precise mathematical treatment of Hintikka’s independence friendly logic. Dependence logic adds the concept of dependence to first order logic. Here the syntax and semantics of dependence logic are studied, dependence logic is given an alternative game theoretic semantics, and sharp results about its complexity are proven. This is a textbook suitable for a special course in logic in mathematics, philosophy, and computer science departments, and contains over 200 exercises, many of which have a full solution at the end of the book. It is also accessible to readers with a basic knowledge of logic, who are interested in new phenomena in logic.

LONDON MATHEMATICAL SOCIETY STUDENT TEXTS Managing editor: Professor J. W. Bruce, Department of Mathematics, University of Hull, UK 3 4 5 8 9 11 12 13 15 17 18 20 21 22 23 24 26 27 28 29 31 32 33 34 35 37 38 39 40 41 42 43 44 45 47 48 49 50 51 52 53 54 55 56 57

Local fields, J. W. S. CASSELS An introduction to twistor theory: Second edition, S. A. HUGGETT & K. P. TOD Introduction to general relativity, L. P. HUGHSTON & K. P. TOD Summing and nuclear norms in Banach space theory, G. J. O. JAMESON Automorphisms of surfaces after Nielsen and Thurston, A. CASSON & S. BLEILER Spacetime and singularities, G. NABER Undergraduate algebraic geometry, MILES REID An introduction to Hankel operators, J. R. PARTINGTON Presentations of groups: Second edition, D. L. JOHNSON Aspects of quantum field theory in curved spacetime, S. A. FULLING Braids and coverings: selected topics, VAGN LUNDSGAARD HANSEN Communication theory, C. M. GOLDIE & R. G. E. PINCH Representations of finite groups of Lie type, FRANCOIS DIGNE & JEAN MICHEL Designs, graphs, codes, and their links, P. J. CAMERON & J. H. VAN LINT Complex algebraic curves, FRANCES KIRWAN Lectures on elliptic curves, J. W. S CASSELS An introduction to the theory of L-functions and Eisenstein series, H. HIDA Hilbert Space: compact operators and the trace theorem, J. R. RETHERFORD Potential theory in the complex plane, T. RANSFORD Undergraduate commutative algebra, M. REID The Laplacian on a Riemannian manifold, S. ROSENBERG Lectures on Lie groups and Lie algebras, R. CARTER, G. SEGAL, & I. MACDONALD A primer of algebraic D-modules, S. C. COUNTINHO Complex algebraic surfaces, A. BEAUVILLE Young tableaux, W. FULTON A mathematical introduction to wavelets, P. WOJTASZCZYK Harmonic maps, loop groups, and integrable systems, M. GUEST Set theory for the working mathematician, K. CIESIELSKI Ergodic theory and dynamical systems, M. POLLICOTT & M. YURI The algorithmic resolution of diophantine equations, N. P. SMART Equilibrium states in ergodic theory, G. KELLER Fourier analysis on finite groups and applications, AUDREY TERRAS Classical invariant theory, PETER J. OLVER Permutation groups, P. J. CAMERON Introductory lectures on rings and modules, J. BEACHY Set theory, A HAJNÁL, P. HAMBURGER K-theory for C*-algebras, M. RORDAM, F. LARSEN, & N. LAUSTSEN A brief guide to algebraic number theory, H. P. F. SWINNERTON-DYER Steps in commutative algebra: Second edition, R. Y. SHARP Finite Markov chains and algorithmic applications, O. HAGGSTROM The prime number theorem, G. J. O. JAMESON Topics in graph automorphisms and reconstruction, J. LAURI & R. SCAPELLATO Elementary number theory, group theory, and Ramanujan graphs, G. DAVIDOFF, P. SARNAK, & A. VALETTE Logic, Induction and Sets, T. FORSTER Introduction to Banach Algebras and Harmonic Analysis, H. G. DALES et al.

58 59 60 61 62 63 64 65 66 67 68 69

Computational Algebraic Geometry, HAL SCHENCK Frobenius Algebras and 2-D Topological Quantum Field Theories, J. KOCK Linear Operators and Linear Systems, J. R. PARTINGTON An Introduction to Noncommutative Noetherian Rings, K. R. GOODEARL & R. B. WARFIELD Topics from One Dimensional Dynamics, K. M. BRUCKS & H. BRUIN Singularities of Plane Curves, C. T. C. WALL A Short Course on Banach Space Theory, N. L. CAROTHERS Elements of the Representation Theory of Associative Algebras Techniques of Representation Theory, IBRAHIM ASSEM, ANDRZEJ SKOWRONSKI, DANIEL SIMSON An Introduction to Sieve Methods and Their Applications, ALINA CARMEN COJOCARU, M. RAM MURTY Elliptic Functions, J. V. ARMITAGE, W. F. EBERLEIN Hyperbolic Geometry from a Local Viewpoint, LINDA KEEN, NIKOLA LAKIC Lectures on Kähler Geometry, ANDREI MOROIANU

LONDON MATHEMATICAL SOCIETY STUDENT TEXTS 70

Dependence Logic A New Approach to Independence Friendly Logic ¨A ¨ NA ¨ NEN J O U KO V A University of Amsterdam and University of Helsinki

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521876599 © J. Vaananen 2007 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2007 eBook (NetLibrary) ISBN-13 978-0-511-28539-4 ISBN-10 0-511-28685-6 eBook (NetLibrary) hardback ISBN-13 978-0-521-87659-9 hardback ISBN-10 0-521-87659-1 paperback ISBN-13 978-0-521-70015-3 paperback ISBN-10 0-521-70015-9 Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Contents

Preface

page ix

1

Introduction

1

2

Preliminaries 2.1 Relations 2.2 Vocabularies and structures 2.3 Terms and formulas 2.4 Truth and satisfaction

5 5 5 6 7

3

Dependence logic 3.1 Examples and a mathematical model for teams 3.2 Formulas as types of teams 3.3 Logical equivalence and duality 3.4 First order formulas 3.5 The flattening technique 3.6 Dependence/independence friendly logic

10 11 16 29 37 42 44

4

Examples 4.1 Even cardinality 4.2 Cardinality 4.3 Completeness 4.4 Well-foundedness 4.5 Connectedness 4.6 Natural numbers 4.7 Real numbers 4.8 Set theory

48 48 51 53 55 56 57 59 60

vii

viii

Contents

5

Game theoretic semantics 5.1 Semantic game of first order logic 5.2 Perfect information game for dependence logic 5.3 Imperfect information game for dependence logic

6

Model theory 6.1 From D to 11 6.2 Applications of 11 6.3 From 11 to D 6.4 Truth definitions 6.5 Model existence game 6.6 Ehrenfeucht–Fra¨ıss´e game for dependence logic

86 86 90 94 100 110 121

7

Complexity 7.1 Decision and other problems 7.2 Some set theory 7.3 2 -completeness in set theory

134 134 135 140

8

Team logic 8.1 Preorder of determination 8.2 Dependence and independence 8.3 Formulas of team logic 8.4 From team logic to L 2 8.5 From L 2 to team logic 8.6 Ehrenfeucht–Fra¨ıss´e game for team logic

144 144 148 150 158 161 163

Appendix: Solutions to selected exercises, by Ville Nurmi References Index

169 220 223

63 63 69 80

Preface

This book is based on lectures I gave at the Department of Mathematics and Statistics, University of Helsinki, during the academic year 2005–2006. I am indebted to the students who followed the course, in particular to Åsa Hirvonen, Meeri Kes¨al¨a, Ville Nurmi, Eero Raaste, and Ryan Siders. Thanks also go to Ville Nurmi for suggesting numerous corrections to the text, compiling the solutions to the exercises in the course, and for allowing me to include the solutions in this book. I am very grateful to Wilfrid Hodges for many useful discussions on dependence. I thank the Newton Institute (Cambridge, UK) for inviting me for the five weeks, during which time the final manuscript was prepared. The preparation of the manuscript was partially supported by grant 40734 of the Academy of Finland. I wish to thank Peter Thompson of Cambridge University Press for all the arrangements concerning publishing, and I am deeply grateful to Juliette Kennedy for her generous help in all stages of writing this book.

ix

1 Introduction

Dependence is a common phenomenon, wherever one looks: ecological systems, astronomy, human history, stock markets. With global warming, the dependence of life on earth on the actions of mankind has become a burning issue. But what is the logic of dependence? In this book we set out to make a systematic logical study of this important concept. Dependence manifests itself in the presence of multitude. A single event cannot manifest dependence, as it may have occurred as a matter of chance. Suppose one day it blows from the west and it rains. There need not be any connection between the wind and the rain, just as if one day it rains and it is Friday the 13th. But over a whole year we may observe that we can tell whether rain is expected by looking at the direction of the wind. Then we would be entitled to say that in the observed location and in the light of the given data, whether it rains depends on the direction of the wind. One would get a more accurate statement about dependence by also observing other factors, such as air pressure. Dependence logic adds the concept of dependence to first order logic. In ordinary first order logic the meaning of the identity x=y

(1.1)

is that the values of x and y are the same. This is a trivial form of dependence. The meaning of fx = y

(1.2)

is that the interpretation of the function symbol f maps the value of x to the value of y. This is an important form of dependence, one where we actually know the mapping which creates the dependence. Note that the dependence 1

2

Introduction

may be more subtle, as in f x z = y. Here y certainly depends on x but also on z. In this case we say that y depends on both x and on z, but is determined by the two together. We introduce the new atomic formulas =(x, y),

(1.3)

the meaning of which is that the values of x and y depend on each other in the particular way that values of x completely determine the values of y. Note the difference between Eqs. (1.1), (1.2) and (1.3). The first says that x determines y in the very strong sense of y being identical with x. The second says that x determines y via the mapping f . Finally, the third says there is some way in which x determines y, but we have no idea what that is. The dependence in Eq. (1.3) is quite common in daily life. We have data that show that weather depends on various factors such as air pressure and air temperature, and we have a good picture of the mathematical equations that these data have to satisfy, but we do not know how to solve these equations, and therefore we do not know how to compute the weather when the critical parameters are given. We could say that the weather obeys dependence of the kind given in Eq. (1.3) rather than of the kind in Eq. (1.2). Historical events typically involve dependencies of the type in Eq. (1.3), as we do not have a perfect theory of history which would explain why events happen the way they do. Human genes undoubtedly determine much of the development of an individual, but we do not know how; we can just see the results. In order to study the logic of dependence we need a framework involving multitude, such as multiple records of historical events, day to day observations of weather and stock transactions. This seems to lead us to study statistics or database theory. These are, however, wrong leads. If we observe that a lamp is lit up four times in a row when we turn a switch, but also that once the lamp does not light up even if we turned the switch (Fig. 1.1), we have to conclude that the light is not completely determined by the switch, as it is by the combined effect of the switch and the plug. From the point of view of dependence, statistical data or a database are relevant only to the extent that they record change. In first order logic the order in which quantifiers are written determines the mutual dependence relations between the variables. For example, in ∀x0 ∃x1 ∀x2 ∃x3 φ the variable x1 depends on x0 , and the variable x3 depends on both x0 and x2 . In dependence logic we write down explicitly the dependence relations

Introduction

3

on

on

on

on

on

off

off

off

off

off

Fig. 1.1. Does the switch determine whether the lamp is lit?

between variables and by so doing make it possible to express dependencies not otherwise expressible in first order logic. The first step in this direction was taken by Henkin with his partially ordered quantifiers, e.g.   ∀x0 ∃x1 φ, ∀x2 ∃x3 where x1 depends only on x0 and x3 depends only on x2 . The remarkable observation about the extension L(H ) of first order logic by this quantifier, made by Ehrenfeucht, was that L(H ) is not axiomatizable. The second step was taken by Hintikka and Sandu, who introduced the slash-notation ∀x0 ∃x1 ∀x2 ∃x3 /∀x0 φ, where ∃x3 /∀x0 means that x3 is “independent” of x0 in the sense that a choice for the value of x3 should not depend on the value of x0 . The observation of Hintikka and Sandu was that we can add slashed quantifiers ∃x3 /∀x0 coherently to first order logic if we give up some of the classical properties of negation, most notably the Law of Excluded Middle. They called their logic independence friendly logic. We take the further step of writing down explicitly the mutual dependence relationships between variables. Thus we write ∀x0 ∃x1 ∀x2 ∃x3 (=(x2 , x3 ) ∧ φ)

(1.4)

to indicate that x3 depends on x2 only. The new atomic formula = (x2 , x3 ) has the explicit meaning that x3 depends on x2 and on nothing else. This results in a logic which we call dependence logic. It is equivalent in expressive power to the logic of Hintikka and Sandu in the sense that there are truth-preserving translations from one to the other. In having the ability to express dependence on the atomic level it is more general.

4

Introduction

Formulas of dependence logic are not like formulas of first order logic. Formulas of dependence logic declare dependencies while formulas of first order logic state relations. These two roles of formulas are incompatible in the following sense. It does not make sense to ask what relation a formula of dependence logic defines, just as it does not make sense to ask what dependence a formula of first order logic states. It seems to the author that the logic of such dependence declarations has not been systematically studied before. At the end of this book we introduce a stronger logic called team logic, reminiscent of the extended independence friendly logic of Hintikka. Team logic is, unlike dependence logic and independence friendly logic, closed under the usual Boolean operations and it satisfies the Law of Excluded Middle.

Historical remarks The possibility of extending first order logic by partially ordered quantifiers was presented by Henkin [14], where also Ehrenfeucht’s result, referred to above, can be found. Independence friendly logic was introduced by Hintikka and Sandu [16] (see also ref. [17]) and advocated strongly by Hintikka in ref. [19]. Hodges [21, 22] gave a compositional semantics for independence friendly logic and we very much follow his approach. Further properties of this semantics are proved in refs. [4], [23] and [41]. Cameron and Hodges [5] showed that there are limitations to the extent to which the semantics can be simplified from the one given in ref. [21]. Connections between independence friendly logic, set theory and second order logic are discussed in ref. [40].

2 Preliminaries

2.1 Relations An n-tuple is a sequence (a1 , . . . , an ) with n components a1 , . . . , an in this order. A special case is the empty sequence ∅, which corresponds to the case n = 0. A relation on a set M is a set R of n-tuples of elements of M for some fixed n, where n is the arity of R. The simplest examples are the usual identity relations on a set M: {(x, x) : x ∈ M}, {(x, x, y) : x, y ∈ M}, {(x, y, x) : x, y ∈ M}, {(x, y, y) : x, y ∈ M}, {(x, x, x) : x ∈ M}. Two special relations are the empty relation ∅, which is the same in any arity, and the unique 0-ary relation {∅}. We think of a function f : M → M as a relation {(x, f (x)) : x ∈ M} on M.

2.2 Vocabularies and structures A vocabulary is a set L of constant, relation and function symbols. We use c to denote constant symbols, R to denote relation symbols, and f to denote function symbols in a vocabulary, possibly with subindexes. Each symbol s in L has an arity # L (s), which is a natural number. The arity of constant symbols is zero. The arity of a relation symbol may be zero. We use x0 , x1 , . . . to denote variables. An L-structure M is a non-empty set M, the domain of M, endowed with an element cM of M for each c ∈ L, an # L (R)-ary relation R M on M for 5

6

Preliminaries

R ∈ L, and an # L ( f )-ary function f M on M for f ∈ L. The L–structures M and M are isomorphic if there is a bijection π : M → M  such that  π(cM ) = cM and for all a1 , . . . , a#L (R) ∈ M we have (a1 , . . . , a#L (R) ) ∈ R M   if and only if (π (a1 ), . . . , π(a#L (R) )) ∈ R M , and f M (π (a1 ), . . . , π(a#L ( f ) )) = π( f M (a1 , . . . , a#L ( f ) )). In this case we say that π is an isomorphism from M to M , denoted π : M ∼ = M . If M is an L-structure and M is an L  -structure such that L  ⊆ L, cM =   M c for c ∈ L  , R M = R M for c ∈ L  , and f M = f M for f ∈ L  , then M  is said to be a reduct of M (to the vocabulary L ), denoted M = ML  , and M is said to be an expansion of M (to the vocabulary L). If M is an L-structure and a ∈ M, then the expansion M of M, denoted (M, a), to a  vocabulary L ∪ {c}, where c ∈ / L, is defined by cM = a; (M, a1 , . . . , an ) is defined similarly.

2.3 Terms and formulas Constant symbols of L and variables are L-terms; if t1 , . . . , tn are L-terms, then f t1 . . . tn is an L-term for each f in L of arity n. The set Var(t) of variables of a term t is simply the set of variables that occur in t. If Var(t) = ∅, then t is called a constant term. For example, f c is a constant term. Every constant term t has a definite value t M in any L-structure M, defined inductively as follows: if t is a constant symbol, t M is defined already. Otherwise, ( f t1 . . . tn )M = f M (t1M , . . . , tnM ). Any function s from a finite set dom(s) of variables into the domain M of an L-structure M is called an assignment of M. Set theoretically, s = {(a, s(a)) : a ∈ dom(s)}. The restriction sA of s to a set A is the function {(a, s(a)) : a ∈ dom(s) ∩ A}. An assignment s assigns a value t M s in M to any L-term t such that Var(t) ⊆ dom(s) as follows: cM s = cM , xnM s = s(xn ), and ( f t1 . . . tn )M s = f M (t1M s, . . . , tnM s). The veritas symbol is an L-formula. Strings ti = t j and Rt1 . . . tn are atomic L-formulas whenever t1 , . . . , tn are L-terms and R is a relation symbol in L with arity n. We sometimes write (ti = t j ) for clarity. Atomic L-formulas are L-formulas. If φ and ψ are L-formulas, then (φ ∨ ψ) and ¬φ are L-formulas. If φ is an L-formula and n ∈ N, then ∃xn φ is an L-formula. We use (φ ∧ ψ) to denote ¬(¬φ ∨ ¬ψ), (φ → ψ) to denote (¬φ ∨ ψ), (φ ↔ ψ) to denote ((φ → ψ) ∧ (ψ → φ)), and ∀xn φ to denote ¬∃xn ¬φ. Formulas defined in this way are called first order. An L-formula is quantifier free if it has no quantifiers.

2.4 Truth and satisfaction

M

7

s

Fig. 2.1. A model and an assignment.

A formula, possibly containing occurrences of the shorthands ∧ and ∀, is in negation normal form if it has negations in front of atomic formulas only. The set Fr(φ) of free variables of a formula φ is defined as follows: Fr(t1 = t2 ) = Var(t1 ) ∪ Var(t2 ), Fr(Rt1 . . . tn ) = Var(t1 ) ∪ . . . ∪ Var(tn ), Fr(φ ∨ ψ) = Fr(φ) ∪ Fr(ψ), Fr(¬φ) = Fr(φ), Fr(∃xn φ) = Fr(φ) \ {xn }. If Fr(φ) = ∅, we call φ an L-sentence.

2.4 Truth and satisfaction Truth in first order logic can be defined in different equivalent ways. The most common approach is the following, based on the more general concept of satisfaction of L-formulas. There is an alternative game theoretic definition of truth, most relevant for this book, and we will introduce it in Chapter 5. In the definition below the concept of an assignment s satisfying an L-formula φ in an L-structure, denoted M |=s φ, is defined by giving a sufficient condition for M |=s φ in terms of subformulas of φ. For quantifiers we introduce the concept of a modified assignment. If s is an assignment and n ∈ N, then s(a/xn ) is the assignment which agrees with s everywhere except that it maps xn to a. In other words, dom(s(a/xn )) = dom(s) ∪ {xn }, s(a/xn )(xi ) = s(xi ) when xi ∈ dom(s) \ {xn }, and s(a/xn )(xn ) = a.

Preliminaries

8

(φ, s, 1)

(φ, s , 0)

Fig. 2.2. Truth and falsity.

We define T as the smallest set such that: (P1) (P2) (P3) (P4) (P5) (P6) (P7) (P8) (P9) (P10)

if t1M s = t2M s, then (t1 = t2 , s, 1) ∈ T ; if t1M s = t2M s, then (t1 = t2 , s, 0) ∈ T ; if (t1M s, . . . , tnM s) ∈ R M , then (Rt1 . . . tn , s, 1) ∈ T ; if (t1M s, . . . , tnM s) ∈ / R M , then (Rt1 . . . tn , s, 0) ∈ T ; if (φ, s, 1) ∈ T or (ψ, s, 1) ∈ T , then (φ ∨ ψ, s, 1) ∈ T ; if (φ, s, 0) ∈ T and (ψ, s, 0) ∈ T , then (φ ∨ ψ, s, 0) ∈ T ; if (φ, s, 1) ∈ T , then (¬φ, s, 0) ∈ T ; if (φ, s, 0) ∈ T , then (¬φ, s, 1) ∈ T ; if (φ, s(a/xn ), 1) ∈ T for some a in M, then (∃xn φ, s, 1) ∈ T ; if (φ, s(a/xn ), 0) ∈ T for all a in M, then (∃xn φ, s, 0) ∈ T .

Finally we define M |=s φ if (φ, s, 1) ∈ T . A formula ψ is said to be a logical consequence of another formula φ, in symbols φ ⇒ ψ, if for all M and s such that M |=s φ we have M |=s ψ. A formula ψ is said to be logically equivalent to another formula φ, in symbols φ ≡ ψ, if φ ⇒ ψ and ψ ⇒ φ. Exercise 2.1 Prove for all first order φ: (φ, s, 1) ∈ T or (φ, s, 0) ∈ T . Exercise 2.2 Prove that for no first order φ and for no s we have (φ, s, 1) ∈ T and (φ, s, 0) ∈ T . Exercise 2.3 Prove only if (φ, s, 1) ∈ /T.

for

all

first

order

φ:

(¬φ, s, 1) ∈ T if and

2.4 Truth and satisfaction

9

We define two operations φ → φ p and φ → φ d by simultaneous induction, using the shorthands φ ∧ ψ and ∀xn φ, as follows: φ d = ¬φ if φ atomic, φ p = φ if φ atomic, (¬φ)d = φ p , (¬φ)p = φ d , (φ ∨ ψ)d = φ d ∧ ψ d , (φ ∨ ψ)p = φ p ∨ ψ p , (∃xn φ)d = ∀xn φ d , (∃xn φ)p = ∃xn φ p . We call φ d the dual of φ. The basic result concerning duality in first order logic is that φ ≡ φ p and ¬φ ≡ φ d . Thus the dual operation is a mechanical way for translating a formula φ to one which is logically equivalent to the negation of φ, without actually adding negation anywhere except in front of atomic formulas. Note that the dual of a formula in negation normal form is again in negation normal form. This is important because negation does not, a priori, preserve the negation normal form, unlike the other logical operations ∧, ∨, ∃, ∀. Exercise 2.4 Show that φ p and φ d are always in negation normal form. d

Exercise 2.5 Prove (φ d ) = φ p and (φ p )p = φ p . p

Exercise 2.6 Compute (φ p )d and (φ d ) . Exercise 2.7 Prove φ ≡ φ p and ¬φ ≡ φ d for any φ in first order logic. Both φ → φ d and φ → φ p preserve logical equivalence. Thus if we define the formula φ ∗ , for any first order formula φ written in negation normal form, to be the result of replacing each logical operation in φ by its dual (i.e. ∧ by ∨ and vice versa, ∀ by ∃ and vice versa), then any logical equivalence φ ≡ ψ gives rise to another logical equivalence φ ∗ ≡ ψ ∗ . This is the Principle of Duality. Exercise 2.8 Prove that φ ≡ ψ implies φ ∗ ≡ ψ ∗ . In terms of game theoretic semantics, which we discuss in Chapter 5, the dual of a sentence, in being logically equivalent to its negation, corresponds to permuting the roles of the players.

3 Dependence logic

Dependence logic introduces the concept of dependence into first order logic by adding a new kind of atomic formula. We call these new atomic formulas atomic dependence formulas. The definition of the semantics for dependence logic is reminiscent of the definition of the semantics for first order logic, presented in Chapter 2. But instead of defining satisfaction for assignments, we follow ref. [21] and jump one level up considering sets of assignments. This leads us to formulate the semantics of dependence logic in terms of the concept of the type of a set of assignments. The reason for the transition to a higher level is, roughly speaking, that one cannot manifest dependence, or independence for that matter, in a single assignment. To see a pattern of dependence, one needs a whole set of assignments. This is because dependence notions can be best investigated in a context involving repeated actions by agents presumably governed by some possibly hidden rules. In such a context dependence is manifested by recurrence, and independence by lack of it. Our framework consists of three components: teams, agents, and features. Teams are sets of agents. Agents are objects with features. Features are like variables which can have any value in a given fixed set. If we have n features and m possible values for each feature, we have altogether m n different agents. Teams are simply subsets of this space of all possible agents. Although our treatment of dependence logic is entirely mathematical, our intuition of dependence phenomena comes from real life examples, thinking of different ways dependence manifests itself in the real world. Statisticians certainly have much to say about this, but when we go deeper into the logic of dependence we see that the crucial concept is determination, not mere 10

3.1 Examples and a mathematical model for teams

11

dependence. Another way in which dependence differs from statistics is that we study total dependence, not statistically significant dependence. It would seem reasonable to define probabilistic dependence logic, but we will not go into that here.

3.1 Examples and a mathematical model for teams In practical examples, a feature is anything that can be in the domain of a function: color, length, weight, prize, profession, salary, gender, etc. To be specific, we use variables, x0 , x1 , . . . to denote features. If features are variables, then agents are assignments. When we define dependence logic, we use the variable xn to refer to the value s(xn ) of the feature xn in an agent s. (i) A team of seven agents with features {x0 , x1 , x2 } as domain and Q as the set of possible values of the features could look like Table 3.1. One can think of this as a set of seven possible assignments to the variables x0 , x1 , x2 in the model (Q, <). Some of the assignments satisfy x0 < x1 and they all satisfy x2 < x1 . (ii) Consider teams of soccer players. In this case the players are the agents. The number assigned to each player as well as the colors of their shirts and shorts are the features, denoted by variables x0 , x1 , x2 , respectively. Teams are sets of players in the usual sense of the word “team.” Table 3.2 depicts a team. If we counted only the color of the players’ shirts and shorts as features, we would obtain the generated team of three agents depicted in Table 3.3. (iii) Databases are good examples of teams. By a database we mean in this context a table of data arranged in columns and rows. The columns are the features, the rows are the agents, and the set consisting of the rows is the Table 3.1. x0 s0 s1 s2 s3 s4 s5 s6

1.5 2.1 2.1 5.1 8.9 21 100

x1

x2

4 4 4 4 4 4 4

0.51 0.55 0.53 0.54 0.53 0.54 0.54

Dependence logic

12

Table 3.2. Soccer players as a team

s0 s1 s2 s3 s4 s5 s6

(Player) x0

(Shirt) x1

(Shorts) x2

1 2 3 4 5 6 7

yellow yellow yellow yellow red red red

white white white white white black black

Table 3.3. A generated team

s0 s1 s2

(Shirt) x1

(Shorts) x2

yellow red red

white white black

Table 3.4. A database Fields Record 1 2 3 ... k

x1

x2

...

xn

52 68 11 ... 101

24 362 7311 ... 43

... ... ... ... ...

1 0 1 ... 1

team. In database theory the columns are often called fields or attributes, and the rows are called records or tuples. Table 3.4 is an example of a database. If the row number (1 to k in Table 3.4) is counted as a feature, then all rows are different agents. Otherwise rows with identical values in all the features are identified, resulting an a team called the generated team, i.e. the team generated by the particular database. Table 3.5 depicts a database and the generated team. (iv) The game history team. Imagine a game where players make moves, following a strategy they have chosen with a certain goal in mind. We are

3.1 Examples and a mathematical model for teams

13

thinking of games in the sense of von Neumann and Morgenstern’s Theory of Games and Economic Behavior [43]. Examples of such games are board and card games, business games, games related to social behavior, etc. We think of the moves of the game as features. If during a game 350 moves are made by the players, then we have 350 features. Plays, i.e. sequences of moves of the game that comprise an entire play of the game, are the agents. Any collection of plays is a team. A team may arise, for example, as follows: two players play a certain game 25 times, thus producing 25 sequences of moves. A team of 25 agents is created. It may be desirable to know answers to the following kinds of questions. (a) What is the strategy that a player is following, or is he or she following any strategy at all? (b) Is a player using information about his or her (or other players’) moves that he or she is committed not to use? This issue is closely related to game theoretic semantics of dependence logic treated in Chapter 5. The following game illustrates how a player can use information that may be not admitted. There are two players I and II. (For ease of reference, we always refer to player I as “he” and player II as “she”.) Player I starts by choosing an integer n. Then II chooses an integer m. After this II makes another move and chooses, this time without seeing n, an integer l. So player II is committed to choose l without seeing n, even if she saw n when she picked m. One may ask, how can she forget a number she has seen once, but if the number has many digits this is quite plausible. Player II wins if l > n. In other words, II has the impossible looking task of choosing an integer l which is bigger than an integer n that she is not allowed to know. Her trick, which we call the signalling-strategy, is to store information about n into m and then choose l only on the basis of what m is. Table 3.5 shows an example of a game history team in this Table 3.5. Game history and the generated team Play 1 2 3 4 5 6 7 8

I

II

II

1 40 2 0 1 2 40 100

1 40 2 0 1 2 40 100

2 41 3 1 2 3 41 101

s0 s1 s2 s3 s4

x0

x1

x2

0 1 2 40 100

0 1 2 40 100

1 2 3 41 101

Dependence logic

14

Table 3.6. Suspicious game history and the generated team Play 1 2 3 4 5 6 7 8

I

II

II

1 40 2 0 1 2 40 100

0 0 0 0 0 0 0 0

2 41 3 1 2 3 41 101

s0 s1 s2 s3 s4

x0

x1

x2

0 1 2 40 100

0 0 0 0 0

1 2 3 41 101

a4 a3 a2 a

x0

x1

Fig. 3.1. Dependence of x1 on x0 .

game. We can see that player II has been using the signalling-strategy. If we instead observed the behavior of Table 3.6, we could doubt whether II is obeying the rules, as her second move seems clearly to depend on the move of I which she is not supposed to see. (v) Every formula φ(x1 , . . . , xn ) of any logic and structure M together give rise to the team of all assignments that satisfy φ(x1 , . . . , xn ) in M. In this case the variables are the features and the assignments are the agents. This (possibly quite large) team manifests the dependence structure φ(x1 , . . . , xn ) that expresses in M. If φ is the first order formula x0 = x1 , then φ expresses the very strong dependence of x1 on x0 , namely of x1 being equal to x0 . The team of assignments satisfying x0 = x1 in a structure is the set of assignments s which give to x0 the same value as to x1 . If φ is the infinitary formula (x0 = x1 ) ∨ (x0 · x0 = x1 ) ∨ (x0 · x0 · x0 = x1 ) ∨ . . . , then φ expresses the dependence of x1 on x0 of being in the set {x0 , x0 · x0 , x0 · x0 · x0 , . . .}. See Fig. 3.1.

3.1 Examples and a mathematical model for teams

15

Fig. 3.2. Picture of a torus team.

(vi) Every first order sentence φ and structure M together give rise to teams consisting of assignments that arise in the semantic game (see Section 5.1) of φ and M. The semantic game is a game for two players, I and II, in which I tries to show that φ is not true in M and II tries to show that φ is indeed true in M. The game proceeds according to the structure of φ. At conjunctions player I chooses a conjunct. At universal quantifiers player I chooses a value for the universally bound variable. At disjunctions player II chooses a disjunct. At existential quantifiers player II picks up a value for the existentially bound variable. At negations the players exchange roles. Thus the players build up move by move an assignment s. When an atomic formula is met, player II wins if the formula is true in M under the assignment s, otherwise player I wins. See Section 5.1 for details. If M |= φ and the winning strategy of II is τ in this semantic game, a particularly interesting team consists of all plays of the semantic game in which II uses τ . This team is interesting because the strategy τ can be read off from the team. We can view the study of teams of plays in this game as a generalization of the study of who wins the semantic game. The semantic game of dependence logic is treated in Chapter 5. (vii) Space team. Let us consider the three-dimensional Euclidean space R3 . Let S be a surface in R3 , e.g. the torus S = {(cos v cos u, cos v sin u, sin v) : u, v ∈ [0, 2π ]}. The set S is a team in which the three coordinates x, y, and z are the features, and the points on the surface are the agents (see Fig. 3.2 and Table 3.7). We now give a mathematical model for teams.

Dependence logic

16

Table 3.7. A torus team x ... 0.2919 0.2829 0.2739 0.2650 0.2563 0.2476 0.2390 0.2305 0.2222 0.2139 0.2057 0.1977 0.1898 0.1820 0.1744 0.1669 0.1595 0.1522 ...

y

z

... 0.4546 0.4504 0.4460 0.4414 0.4366 0.4316 0.4265 0.4212 0.4157 0.4101 0.4042 0.3983 0.3922 0.3859 0.3794 0.3729 0.3661 0.3592 ...

... 0.8415 0.8468 0.8521 0.8573 0.8624 0.8674 0.8724 0.8772 0.8820 0.8866 0.8912 0.8957 0.9001 0.9044 0.9086 0.9128 0.9168 0.9208 ...

Definition 3.1 An agent is any function s from a finite set dom(s) of variables, also called features, to a fixed set M. The set dom(s) is called the domain of s, and the set M is called the codomain of s. A team is any set X of agents with the same domain, called the domain of X and denoted by dom(X ), and the same codomain, likewise called the codomain of X . A team with codomain M is called a team of M. If V is a finite set of variables, we use Team(M, V ) to denote the set of all teams of M with domain V . Since we have defined teams as sets, not multisets, of assignments, one assignment can occur only once in a team. Allowing multisets would not, however, change anything essential in this study.

3.2 Formulas as types of teams We define a logic which has an atomic formula for expressing dependence. We call this logic the dependence logic and denote it by D. In Section 3.6 we will recover independence friendly logic as a fragment of dependence logic.

3.2 Formulas as types of teams

17

M X

Fig. 3.3. A model and a team.

In first order logic the meaning of a formula is derived from the concept of an assignment satisfying the formula. In dependence logic the meaning of a formula is based on the concept of a team being of the (dependence) type of the formula. Recall that teams are sets of agents (assignments) and that agents are functions from a finite set of variables, called the domain of the agent into an arbitrary set called the codomain of the agent (Definition 3.1). In a team the domain of all agents is assumed to be the same set of variables, just as the codomain of all agents is assumed to be the same set ( Fig. 3.3). Our atomic dependence formulas have the following form: =(t1 , . . . , tn ). The intuitive meaning of this is as follows: the value of the term tn depends only on the values of the terms t1 , . . . , tn−1 .

As singular cases, we have =(), which we take to be universally true, and =(t), which declares that the value of the term t depends on nothing, i.e. is constant. Note that =(x1 ) is quite non-trivial and indispensable if we want to say that all agents are similar as far as feature x1 is concerned. Such similarity is manifested by the team of Table 3.6, where all agents have value 0 in their feature x1 . Actually, our atomic formulas express determination rather than dependence. The reason for this is that determination is a more basic

18

Dependence logic

concept than dependence. Once we can express determination, we can define dependence and independence. Already dependence logic has considerable strength. Further extensions formalizing the concepts of dependence and independence are even stronger, and in addition lack many of the nice model theoretic properties that our dependence logic enjoys. We will revisit the concepts of dependence and particularly independence in Section 8.2 Definition 3.2 Suppose L is a vocabulary. If t1 , . . . , tn are L-terms and R is a relation symbol in L with arity n, then strings ti = t j , =(t1 , . . . , tn ), Rt1 . . . tn are L-formulas of dependence logic D. They are called atomic formulas. If φ and ψ are L-formulas of D, then (φ ∨ ψ) and ¬φ are L-formulas of D. If φ is an L-formula of D and n ∈ N, then ∃xn φ is an L-formula of D. As is apparent from Definition 3.2, the syntax of dependence logic D is very similar to that of first order logic, the only difference being the inclusion of the new atomic formulas =(t1 , . . . , tn ). We use (φ ∧ ψ) to denote ¬(¬φ ∨ ¬ψ), (φ → ψ) to denote (¬φ ∨ ψ), (φ ↔ ψ) to denote ((φ → ψ) ∧ (ψ → φ)), and ∀xn φ

3.2 Formulas as types of teams

19

X(F/xn )

Fig. 3.4. A supplement team.

to denote ¬∃xn ¬φ. A formula of dependence logic which does not contain any atomic formulas of the form =(t1 , . . . , tn ) is called first order. The veritas symbol is definable as =(). We call this also first order as we took as a special symbol in Chapter 2. The set Fr(φ) of free variables of a formula φ is defined otherwise as for first order logic, except that we have the new case Fr(=(t1 , . . . , tn )) = Var(t1 ) ∪ · · · ∪ Var(tn ). If Fr(φ) = ∅, we call φ an L-sentence of dependence logic. We define now two important operations on teams, the supplement and the duplication operations. The supplement operation adds a new feature to the agents in a team, or alternatively changes the value of an existing feature. Suppose a strategy officer of a company comes to a director with a plan for a team to design a new product. The director asks: “What about the language skills of the team members?” The strategy officer answers: “No problem, I can supplement the language skills to the team description, and then you will see that the team is really of the type we need.” This is the idea of supplementing a team (see Fig. 3.4). Definition 3.3 If M is a set, X is a team with M as its codomain and F : X → M, we let X (F/xn ) denote the supplement team {s(F(s)/xn ) : s ∈ X }. A duplicate team is obtained by duplicating the agents of a team until all possibilities occur as far as a particular feature is concerned. Suppose a strategy officer of a company comes to a director with a plan for a team for a company wide committee. The strategy officer says, “I decided we need a programmer, an analyst, and a sales person. I chose such people from

20

Dependence logic

X (M/ xn)

Fig. 3.5. A duplicate team.

each of our five departments.” This is the idea of duplicating a team. The team to be duplicated consisted of three agents with just one feature with values in the set programmer, analyst, sales person. The duplicated team has 15 agents corresponding to a programmer, an analyst, and a sales person from each of the five departments. As this example indicates, in real life examples the features do not always have values in the same set, as in our mathematical model. We could rectify this by considering many-sorted structures, but this would lead to unnecessarily complicated notation. Definition 3.4 If M is a set and X is a team of M we use X (M/xn ) to denote the duplicate team {s(a/xn ) : a ∈ M, s ∈ X }. See Fig. 3.5. We are ready to define the semantics of dependence logic. Definition 3.5 Suppose L is a vocabulary and M is an L-structure. We define the set T , or more exactly TM , called the fundamental predicate of M, of triples (φ, X, d), where φ is an L-formula of dependence logic, Fr(φ) ⊆ dom(X ) and d ∈ {0, 1}, as the smallest set such that (D1) if t1M s = t2M s for all s ∈ X , then (t1 = t2 , X, 1) ∈ T ; (D2) if t1M s = t2M s for all s ∈ X , then (t1 = t2 , X, 0) ∈ T ; (D3) if tnM s = tnM s   for all s, s  ∈ X such that M M t1M s = t1M s  , . . . , tn−1

s = tn−1

s  ,

then (=(t1 , . . . , tn ), X, 1)) ∈ T ; (D4) (=(t1 , . . . , tn ), ∅, 0) ∈ T ; (D5) if (t1M s, . . . , tnM s) ∈ R M for all s ∈ X , then (Rt1 . . . tn , X, 1) ∈ T ; (D6) if (t1M s, . . . , tnM s) ∈ / R M for all s ∈ X , then (Rt1 . . . tn , X, 0) ∈ T ;

3.2 Formulas as types of teams

21

(φ, X, 1)

(φ, X, 0)

Fig. 3.6. Truth and falsity.

(D7) if (φ, X, 1) ∈ T , (ψ, Y, 1) ∈ T , and dom(X ) = dom(Y ), then (φ ∨ ψ, X ∪ Y, 1) ∈ T ; (D8) if (φ, X, 0) ∈ T and (ψ, X, 0) ∈ T , then (φ ∨ ψ, X, 0) ∈ T ; (D9) if (φ, X, 1) ∈ T , then (¬φ, X, 0) ∈ T ; (D10) if (φ, X, 0) ∈ T , then (¬φ, X, 1) ∈ T ; (D11) if (φ, X (F/xn ), 1) ∈ T for some F : X → M, then (∃xn φ, X, 1) ∈ T ; (D12) if (φ, X (M/xn ), 0) ∈ T , then (∃xn φ, X, 0) ∈ T . Finally, we define X is of type φ in M, denoted M |= X φ if (φ, X, 1) ∈ T . Furthermore, φ is true in M, denoted M |= φ, if M |={∅} φ, and φ is valid, denoted |= φ, if M |= φ for all M. Note that, M |= X ¬φ if (φ, X, 0) ∈ T ; M |= ¬φ if (φ, {∅}, 0) ∈ T . Then we say that φ is false in M. We will see in a moment that it is not true in general that (φ, X, 1) ∈ T or (φ, X, 0) ∈ T . Likewise it is not true in general that M |= φ or M |= ¬φ, nor that M |= φ ∨ ¬φ. In other words, no sentence can be both true and false in a model, but some sentences can be neither true nor false in a model. This gives our logic a nonclassical flavor. See Fig. 3.6.

Dependence logic

22

Table 3.8.

s0 s1 s2

x0

x1

x2

x3

0 1 0

0 0 0

0 1 0

1 0 1

Table 3.9.

s0 s1 s2

x0

x1

x2

x3

red white black

yellow red red

white white black

white white white

Example 3.6 Let M be a structure with M = {0, 1}. Consider the team given in Table 3.8. This team is of type =(x1 ), since si (x1 ) = 0 for all i. This team is of type x0 = x2 , as si (x0 ) = si (x2 ) for all i. This team is of type ¬x0 = x3 , as si (x0 ) = si (x3 ) for all i. This team is of type =(x0 , x1 ), as si (x0 ) = s j (x0 ) implies si (x3 ) = s j (x3 ). This team is not of type =(x1 , x2 ), as s0 (x1 ) = s1 (x1 ), but s0 (x2 ) = s1 (x2 ). Finally, this team is of type =(x0 ) ∨ =(x0 ), as it can be represented as the union {s0 , s2 } ∪ {s1 }, where {s0 , s2 } and {s1 } are both of type =(x0 ). Example 3.7 Let M be a structure with M = {yellow, white, r ed, black}. Consider the team in Table 3.9. This team is of type =(x3 ), since si (x3 ) = white for all i. This team is also of type =(x0 , x1 ), as si (x0 ) = s j (x0 ) implies i = j and hence si (x1 ) = s j (x1 ). This team is not of type =(x1 , x2 ), as s1 (x1 ) = s2 (x1 ), but s1 (x2 ) = s2 (x2 ). Definition 3.5 refers to a set T of triples. Such a smallest set T always exists. For example, the set of all possible triples (φ, X, d) satisfies (D1)–(D12), and the intersection of all sets T with (D1)–(D12) has again the properties (D1)– (D12). We may also construct the minimal T step by step starting from T0 = ∅ and letting Tn+1 be an extension of Tn satisfying the consequents if Tn satisfies the antecedents of (D1)–(D12). Then the minimal T is ∪n Tn . Proposition 8 shows that the implications of Definition 3.5 can all be reversed, making T a fixed point of the inductive definition of truth.

3.2 Formulas as types of teams

23

Proposition 3.8 The set T satisfies: (E1) (t1 = t2 , X, 1) ∈ T if and only if for all s ∈ X we have t1M s = t2M s; (E2) (t1 = t2 , X, 0) ∈ T if and only if for all s ∈ X we have t1M s = t2M s; (E3) (=(t1 , . . . , tn ), X, 1) ∈ T if and only if for all s, s  ∈ X such that M M t1M s = t1M s  , . . . ,tn−1

s = tn−1

s  , we have tnM s = tnM s  ; (E4) (=(t1 , . . . , tn ), X, 0) ∈ T if and only if X = ∅; (E5) (Rt1 . . . tn , X, 1) ∈ T if and only if for all s ∈ X we have (t1M s, . . . , tnM s) ∈ R M ; (E6) (Rt1 . . . tn , X, 0) ∈ T if and only if for all s ∈ X we have (t1M s, . . . , tnM s) ∈ / RM; (E7) (φ ∨ ψ, X, 1) ∈ T if and only if X = Y ∪ Z such that dom(Y ) = dom(Z ), (φ, Y, 1) ∈ T and (ψ, Z , 1) ∈ T ; (E8) (φ ∨ ψ, X, 0) ∈ T if and only if (φ, X, 0) ∈ T and (ψ, X, 0) ∈ T ; (E9) (¬φ, X, 0) ∈ T if and only if (φ, X, 1) ∈ T ; (E10) (¬φ, X, 1) ∈ T if and only if (φ, X, 0) ∈ T ; (E11) (∃xn φ, X, 1) ∈ T if and only if (φ, X (F/xn ), 1) ∈ T for some F : X → M; (E12) (∃xn φ, X, 0) ∈ T if and only if (φ, X (M/xn ), 0) ∈ T . Proof Suppose (θ, X, d) is a triple for which one of the claims (E1)–(E12) fails. Note that we can read uniquely from the triple whose claim it is that fails. Let T  = T \ {(θ, X, d)}. We show that T  satisfies (D1)–(D12), which contradicts the minimality of T . We leave as an exercise the fact that T  satisfies (D1)– (D6). We can see that T  satisfies (D7): if (φ, X, 1) ∈ T  and (ψ, Y, 1) ∈ T  , then ((φ ∨ ψ), X ∪ Y, 1) ∈ T  , unless ((φ ∨ ψ), X, 1) = (θ, X, 1). Since θ is a disjunction, (E7) in such a case fails. Thus (φ, X, 1) ∈ / T or (ψ, Y, 1) ∈ / T . But   this contradicts the assumption that (φ, X, 1) ∈ T and (ψ, Y, 1) ∈ T . Also T  satisfies (D8): if (φ, X, 0) ∈ T  and (ψ, X, 0) ∈ T  , then ((φ ∨ ψ), X, 0) ∈ T  , unless ((φ ∨ ψ), X, 0) = (θ, X, d). Since θ is a disjunction, (E8) in such a case fails. Thus (φ, X, 0) ∈ / T or (ψ, X, 0) ∈ / T . But this contradicts the assump tion that (φ, X, 0) ∈ T and (ψ, X, 0) ∈ T  . We leave the other cases as an exercise.  Note that r (φ ∧ ψ, X, 1) ∈ T if and only if (φ, X, 1) ∈ T and (ψ, X, 1) ∈ T ; r (φ ∧ ψ, X, 0) ∈ T if and only if X = Y ∪ Z such that dom(Y ) = dom(Z ), (φ, Y, 0) ∈ T , and (ψ, Z , 0) ∈ T ; r (∀x φ, X, 1) ∈ T , if and only if (φ, X (M/x ), 1) ∈ T ; n n r (∀x φ, X, 0) ∈ T if and only if (φ, X (F/x ), 0) ∈ T for some F : X → M. n n

24

Dependence logic

It may seem strange to define (D4) as (=(t1 , . . . , tn ), ∅, 0) ∈ T . Why not allow (=(t1 , . . . , tn ), X, 0) ∈ T for non-empty X ? The reason is that if we M M negate “for all s, s  ∈ X such that t1M s = t1M s  , . . . , tn−1

s = tn−1

s  , we M M  have tn s = tn s ,” maintaining the analogy with (D2) and (D6), we get M M “for all s, s  ∈ X we have t1M s = t1M s  , . . . , tn−1

s = tn−1

s   and tnM s = M  tn s ,” which is only possible if X = ∅. Some immediate observations can be made using Definition 3.5. We first note that the empty team ∅ is of the type of any formula, as (φ, ∅, 1) ∈ T holds for all φ. In fact, we have the following lemma. Lemma 3.9 For all φ and M we have (φ, ∅, 1) ∈ T and (φ, ∅, 0) ∈ T . Proof Inspection of Definition 3.5 reveals that all the necessary implications hold vacuously when X = ∅.  In other words, the empty team is for all φ of type φ and of type ¬φ. Since the type of a team is defined by reference to all agents in the team, the empty team ends up having all types, just as it is usually agreed that the intersection of an empty collection of subsets of a set M is the set M itself. A consequence of this is that there are no formulas φ and ψ of dependence logic such that M |= X φ implies M |= X ψ, for all M and all X . Namely, letting X = ∅ would yield a contradiction. The following test is very important and will be used repeatedly in the sequel. Closure downwards is a fundamental property of types in dependence logic. Proposition 3.10 (Closure Test) Suppose Y ⊆ X . Then M |= X φ implies M |=Y φ. Proof Every condition from (E1) to (E12) in Proposition 3.8 is closed under taking a subset of X . So if (φ, X, 1) ∈ T and Y ⊆ X , then (φ, Y, 1) ∈ T .  The intuition behind the Closure Test is as follows. To witness the failure of dependence we need a counterexample, two or more assignments that manifest the failure. The smaller the team, the fewer the number of counterexamples. In a one-agent team, a counterexample to dependence is no longer possible. On the other hand, the bigger the team, the more likely it is that some lack of dependence becomes exposed. In the maximal team of all possible assignments no dependence is possible, unless the universe has just one element. Corollary 3.11 There is no formula φ of dependence logic such that for all X = ∅ and all M we have M |= X φ ⇐⇒ M |= X =(x0 , x1 ).

3.2 Formulas as types of teams

25

Proof Suppose for a contradiction M has at least two elements a, b. Let X consist of s = {(x0 , a), (x1 , a)} and s  = {(x0 , a), (x1 , b)}. Now M |= X =(x0 , x1 ), so M |= X φ. By the Closure Test, M |={s} φ, whence M |={s} =(x0 , x1 ). But this is clearly false.  We can replace “all M” by “some M with more than one element in the universe” in Corollary 3.11. Note that in particular we do not have for all X = ∅: M |= X ¬ =(x0 , x1 ) ⇐⇒ M |= X =(x0 , x1 ). Example 3.12 Every team X , the domain of which contains xi and x j , is of type xi = x j ∨ ¬xi = x j , as we can write X = Y ∪ Z , where Y = {s ∈ X : s(xi ) = s(x j )} and Z = {s ∈ X : s(xi ) = s(x j )}. Note that then Y is of type xi = x j , and Z is of type xi = x j . The following example formalizes the idea that we can always choose x1 such that it only depends on x0 , if there are no other requirements. The most obvious way to establish the dependence is to choose x1 identical to x0 . Example 3.13 |= ∀x0 ∃x1 (=(x0 , x1 )). To prove this, fix M. Let X be the set of s : {x0 , x1 } → M with s(x1 ) = s(x0 ). We use Definition 3.5. By (D3), (=(x0 , x1 ), X, 1) ∈ T , for if s, s  ∈ X and s(x0 ) = s  (x0 ), then s(x1 ) = s  (x1 ). Let Y be the set of s : {x0 } → M. If we let F(s) = s(x0 ) then by (D11) (∃x1 (=(x0 , x1 )), Y, 1) ∈ T . Thus (∀x0 ∃x1 (=(x0 , x1 )), {∅}, 1) ∈ T . In Example 3.14 there is another variable in the picture, but it does not matter. Example 3.14 |= ∀x0 ∀x2 ∃x1 (=(x0 , x1 )). To prove this, fix M. Let X be the set of assignments s : {x0 , x1 , x2 } → M with s(x1 ) = s(x0 ). We use Definition 3.5. By (D3), (=(x0 , x1 ), X, 1) ∈ T . Let Y be the set of all s : {x0 , x1 } → M. If we let F(s) = s(x0 ), by (D11) (∃x1 (=(x0 , x1 )), Y, 1) ∈ T . Thus the triple (∀x0 ∀x2 ∃x1 (=(x0 , x1 )), {∅}, 1) is in T . The following example confirms the intuition that functions convey dependence. If a model has a function f M , then we can pick for every a ∈ M an element b ∈ M which depends only on a via the function f M . Example 3.15 Let f be a function symbol of the vocabulary. Then always |= ∀x0 ∃x1 (=(x0 , x1 ) ∧ x1 = f x0 ). To prove this, fix M. Let X be the set of s : {x0 , x1 } → M with s(x1 ) = f M (s(x0 )). We use Definition 3.5. By (D1), ((x1 = f x0 ), X, 1) ∈ T . By (D3), (=(x0 , x1 ), X, 1) ∈ T . Thus by (D7), ((=(x0 , x1 ) ∧ x1 = f x0 ), X, 1) ∈ T . Let Y be the set of s : {x0 } → M. If we let F(s) = f M (s(x0 )) we get from (D11) (∃x1 (=(x0 , x1 ) ∧ x1 = f x0 ), Y, 1) ∈ T . Thus (∀x0 ∃x1 (=(x0 , x1 ) ∧ x1 = f x0 ), {∅}, 1) ∈ T .

26

Dependence logic

The following example is another way of saying that the identity x0 = x1 implies the dependence of x1 on x0 . This is the most trivial kind of dependence. Example 3.16 |= ∀x0 ∀x1 (x0 = x1 → =(x0 , x1 )). To prove this, fix M. Let X be the set of all assignments s : {x0 , x1 } → M. Let Y = {s ∈ X : s(x0 ) = s(x1 )} and Z = X \ Y . Now (x0 = x1 , Z , 0) ∈ T , i.e. (¬(x0 = x1 ), Z , 1) ∈ T and (=(x0 , x1 ), Y, 1) ∈ T . Thus, ((x0 = x1 → =(x0 , x1 )), X, 1) ∈ T . It is important to take note of a difference between universal quantification in first order logic and universal quantification in dependence logic. It is perfectly possible to have a formula φ(x0 ) of dependence logic of the empty vocabulary with just x0 free such that for a new constant symbol c we have |= φ(c) and still |= ∀x0 φ(x0 ), as the following example shows. For this example, remember that =(x1 ) is the type “x1 is constant” of a team in which all agents have the same value for their feature x1 . Recall also the definition of the expansion (M, a) in Section 2.2. Example 3.17 Suppose M is a model of the empty1 vocabulary with at least two elements in its domain. Let φ be the sentence ∃x1 (=(x1 ) ∧ c = x1 ) of dependence logic. Then (M, a) |= ∃x1 (=(x1 ) ∧ c = x1 )

(3.1)

for all expansions of (M, a) of M to the vocabulary {c}. To prove Eq. (3.1) suppose we are given an element a ∈ M. We can define Fa (∅) = a and then (M, a) |={{(x1 ,a)}} (=(x1 ) ∧ c = x1 ), where we have used {∅}(Fa /x1 ) = {{(x1 , a)}}. However, M |= ∀x0 ∃x1 (=(x1 ) ∧ x0 = x1 ).

(3.2)

To prove Eq. (3.2) suppose the contrary, that is M |={∅} ∀x0 ∃x1 (=(x1 ) ∧ x0 = x1 ). Then M |={{(x0 ,a)}:a∈M} ∃x1 (=(x1 ) ∧ x0 = x1 ), 1

The empty vocabulary has no constant, relation or function symbols. Structures for the empty vocabulary consist of merely a non-empty set as the universe.

3.2 Formulas as types of teams

27

where we have written {∅}(M/x0 ) out as {{(x0 , a)} : a ∈ M}. Let F : {{(x0 , a)} : a ∈ M} → M such that M |={{(x0 ,a),(x1 ,G(a))}:a∈M} (=(x1 ) ∧ x0 = x1 ),

(3.3)

where G(a) = F({(x0 , a)}) and {{(x0 , a)} : a ∈ M}(F/x1 ) has been written as {{(x0 , a), (x1 , G(a))} : a ∈ M}. In particular, M |={{(x0 ,a),(x1 ,G(a))}:a∈M} =(x1 ), which means that F has to have a constant value. Since M has at least two elements, the consequence M |={{(x0 ,a),(x1 ,G(a))}:a∈M} x0 = x1 of Eq. (3.3) contradicts (D1). Example 3.18 A team X is of type ∃x0 (=(x2 , x0 ) ∧ x0 = x1 ) if and only if every s ∈ X can be modified to s(as /x0 ) such that for all s ∈ X (i) as = s(x1 ), (ii) as depends only on s(x2 ) in X , if and only if X is of type =(x2 , x1 ). Example 3.19 A team X is of type ∃x0 (= (x2 , x0 ) ∧ ¬(x0 = x1 )) if every s ∈ X can be modified to s(as /x0 ) in such a way that (i) as = s(x1 ), (ii) as is dependent only on s(x2 ) in X . This means that we have to be able to determine what s(x1 ) is, in order to avoid it, only on the basis of what s(x2 ) is. (See also Exercise 3.15.) Example 3.20 A team X with domain {x1 } is of type ∃x0 (=(x0 ) ∧ Rx0 x1 ) if every s ∈ X , which is really just s(x1 ), can be modified to s(as /x0 ) such that s(as /x0 ) satisfies Rx0 x1 and as is the same for all s ∈ X . That is, there is one a such that {s(a/x0 ) : s ∈ X } is of type Rx0 x1 . Exercise 3.1 Suppose L = {R}, # L (R) = 2. Show that every team X , the domain of which contains xi and x j , is of type Rxi x j ∨ ¬Rxi x j . Exercise 3.2 Describe teams X ∈ Team(M, {x0 , x1 }) of type (i) =(x0 , x1 ), (ii) =(x1 , x0 ), (iii) =(x0 , x0 ).

Dependence logic

28

Exercise 3.3 Let L be the vocabulary {c, f }. Depending on the model M, which teams X ∈ Team(M, {x0 }) are of the following types? (i) (ii) (iii) (iv)

=(c, c), =(x0 , c), =(c, x0 ), =(c, f x0 ).

Exercise 3.4 Let M = (N, +, ·, 0, 1). Which teams X ∈ Team(M, {x0 , x1 }) are of the following types? (i) =(x0 , x0 + x1 ), (ii) =(x0 · x0 , x1 · x1 ). Exercise 3.5 Let L be the vocabulary { f, g}. Which teams Team(M, {x0 , x1 }) are of the following types?

X∈

(i) =( f x0 , x0 ), (ii) =( f x1 , x0 ), (iii) =( f x0 , gx1 ). Exercise 3.6 Let L be the vocabulary { f, g}. Describe teams X ∈ Team(M, {x0 , x1 , x2 }) of type (i) =(x0 , x1 , x2 ), (ii) =(x0 , x0 , x2 ). Exercise 3.7 Let M = (N, +, ·, 0, 1) and X n = {{(0, a), (1, b)} : 1 < a ≤ n, 1 < b ≤ n, a ≤ b}. Show that X 5 is of type =(x0 + x1 , x0 · x1 , x0 ). This is also true for X n for any n, but is slightly harder to prove. Exercise 3.8 |= ∀x1 . . . ∀xn (=(x1 , . . . , xn , xi )), if 1 ≤ i ≤ n. Exercise 3.9 |= ∀x0 ∀x1 (x1 = c → =(x0 , x1 )). Exercise 3.10 |= ∀x0 ∀x1 (x1 = f x0 → =(x0 , x1 )). Exercise 3.11 For which of the following formulas φ is it true that for all X = ∅: M |= X ¬φ ⇐⇒ M | = X φ: (i) (=(x0 , x1 ) ∧ ¬x0 = x1 ), (ii) (=(x0 , x1 ) → x0 = x1 ), (iii) (=(x0 , x1 ) ∨ ¬x0 = x1 )?

3.3 Logical equivalence and duality

29

Exercise 3.12 Show that the set T of all triples (φ, X, d), where φ is an L-formula of dependence logic, Fr(φ) ⊆ dom(X ) and d ∈ {0, 1}, satisfies (D1)– (D12) of Definition 3.5. Exercise 3.13 Show that the family of all sets T satisfying (D1)–(D12) of Definition 3.5 is closed under arbitrary intersections. Exercise 3.14 Finish the proof of Proposition 3.8. Exercise 3.15 The following appealing claim is wrong. Why? Suppose M is an L-structure with ≥ 2 elements. Then a team X is of type ∃x0 (=(x2 , x0 ) ∧ x0 = x1 ) if and only if it is of type ∃x0 (=(x2 , x0 ) ∧ ¬(x0 = x1 )). Exercise 3.16 The following appealing claim is wrong. Why? Suppose M is an L-structure with ≥ 2 elements. Then a team X is of type ∃x0 (=(x0 ) ∧ x0 = x1 ) if and only if it is of type ∃x0 (=(x0 ) ∧ ¬(x0 = x1 )). Exercise 3.17 The following appealing claim is wrong. Why? Suppose L = { f }, #( f ) = 1, and M is an L-structure. Then a team with domain {x0 } is of type ∃x1 (=( f x0 , x1 ) ∧ f x1 = f x0 ) if and only if the mapping s(x0 ) → f M (s(x0 )) is one-to-one for s ∈ X . Exercise 3.18 The following appealing claim is wrong. Why? Suppose L = {R}, #(R) = 2, and M is an L-structure. Then a team X with domain {x0 }, which we identify with a subset of M, is of type ∃x1 (=(x0 , x1 ) ∧ Rx0 x1 ) if and only if R M is a function with domain X . Exercise 3.19 ([21]) This exercise shows that the Closure Test is the best we can do. Let L be the vocabulary of one n-ary predicate symbol R. Let M be a finite set and m ∈ N. Suppose S is a set of teams of M with domain {x1 , . . . , xm } such that S is closed under subsets. Find an interpretation R M ⊆ M n and a formula φ of D such that a team X with domain {x1 , . . . , xk } is of type φ in M if and only if X ∈ S. Exercise 3.20 Use the method of ref. [5], mutatis mutandis, to show that there is no compositional semantics for dependence logic in which the meanings of formulas are sets of assignments (rather than sets of teams) and which agrees with Definition 3.5 for sentences.

3.3 Logical equivalence and duality The concept of logical consequence and the derived concept of logical equivalence are both defined below in a semantic form. In first order logic there is also

Dependence logic

30

(φ, X, 1)

(ψ, X, 1)

Fig. 3.7. Logical consequence φ ⇒ ψ.

a proof theoretic (or syntactic) concept of logical consequence and it coincides with the semantic concept. This fact is referred to as the G¨odel Completeness Theorem. In dependence logic we have only semantic notions. There are obvious candidates for syntactic concepts but they are not well understood yet. For example, it is known that the G¨odel Completeness Theorem fails badly (see Section 4.6). Definition 3.21 ψ is a logical consequence of φ, φ ⇒ ψ, if for all M and all X with dom(X ) ⊇ Fr(φ) ∪ Fr(ψ) and M |= X φ we have M |= X ψ. Further, ψ is a strong logical consequence of φ, φ ⇒∗ ψ, if for all M and for all X with dom(X ) ⊇ Fr(φ) ∪ Fr(ψ) and M |= X φ we have M |= X ψ, and all X with dom(X ) ⊇ Fr(φ) ∪ Fr(ψ) and M |= X ¬ψ we have M |= X ¬φ. Also, ψ is logically equivalent with φ, φ ≡ ψ, if φ ⇒ ψ and ψ ⇒ φ. Lastly, ψ is strongly logically equivalent with φ, φ ≡∗ ψ, if φ ⇒∗ ψ and ψ ⇒∗ φ. See Figs. 3.7 and 3.8. Note that φ ⇒∗ ψ if and only if φ ⇒ ψ and ¬ψ ⇒ ¬φ. Thus the fundamental concept is φ ⇒ ψ and φ ⇒∗ ψ reduces to it. Note also that φ and ψ are logically equivalent if and only if for all X with dom(X ) ⊇ Fr(φ) ∪ Fr(ψ) (φ, X, 1) ∈ T if and only if (ψ, X, 1) ∈ T ,

3.3 Logical equivalence and duality

31

(φ, Y, 0) (φ, X, 1) (ψ, Y, 0) (ψ, X, 1)

Fig. 3.8. Strong logical consequence φ ⇒∗ ψ.

and φ and ψ are strongly logically equivalent if and only if for all X with dom(X ) ⊇ Fr(φ) ∪ Fr(ψ) and all d, (φ, X, d) ∈ T if and only if (ψ, X, d) ∈ T . We have some familiar looking strong logical equivalences in propositional logic, reminiscent of axioms of semigroups with identity. In Lemma 3.22 we group the equivalences according to duality. Lemma 3.22 The following strong logical equivalences hold in dependence logic: (i) ¬¬φ ≡∗ φ; (ii) (a) (φ ∧ ) ≡∗ φ, (b) (φ ∨ ) ≡∗ ; (iii) (a) (φ ∧ ψ) ≡∗ (ψ ∧ φ), (b) (φ ∨ ψ) ≡∗ (ψ ∨ φ), (iv) (a) (φ ∧ ψ) ∧ θ ≡∗ φ ∧ (ψ ∧ θ), (b) (φ ∨ ψ) ∨ θ ≡∗ φ ∨ (ψ ∨ θ); (v) (a) ¬(φ ∨ ψ) ≡∗ (¬φ ∧ ¬ψ), (b) ¬(φ ∧ ψ) ≡∗ (¬φ ∨ ¬ψ). Proof We prove only Claim (iii) (b) and leave the rest to the reader. By (E8), (φ ∨ ψ, X, 0) ∈ T if and only if ((φ, X, 0) ∈ T and (ψ, X, 0) ∈ T ) if and only if (ψ ∨ φ, X, 0) ∈ T . Suppose then (φ ∨ ψ, X, 1) ∈ T . By (E7) there are Y and Z such that X = Y ∪ Z , (φ, Y, 1) ∈ T , and (ψ, Z , 1) ∈ T . By (D7), (ψ ∨ φ, X, 1) ∈ T . Conversely, if (ψ ∨ φ, X, 1) ∈ T , then there are Y and Z such that X = Y ∪ Z , (ψ, Y, 1) ∈ T , and (φ, Z , 1) ∈ T . By (D7), (φ ∨ ψ, X, 1) ∈T. 

Dependence logic

32

Many familiar propositional equivalences fail on the level of strong equivalence, in particular the Law of Excluded Middle, weakening laws, and distributivity laws. See Exercise 3.21. We also have some familiar looking strong logical equivalences for quantifiers. In Lemma 3.23 we again group the equivalences according to duality. Lemma 3.23 The following strong logical equivalences and consequences hold in dependence logic: (i) (a) (b) (ii) (a) (b) (iii) (a) (b) (iv) (a) (b)

∃xm ∃xn φ ≡∗ ∃xn ∃xm φ, ∀xm ∀xn φ ≡∗ ∀xn ∀xm φ; ∃xn (φ ∨ ψ) ≡∗ (∃xn φ ∨ ∃xn φ), ∀xn (φ ∧ ψ) ≡∗ (∀xn φ ∧ ∀xn φ); ¬∃xn φ ≡∗ ∀xn ¬φ, ¬∀xn φ ≡∗ ∃xn ¬φ; φ ⇒∗ ∃xn φ, ∀xn φ ⇒∗ φ.

A useful method for proving logical equivalences is the method of substitution. In first order logic this is based on the strong compositionality2 of the semantics. The same is true for dependence logic. Definition 3.24 Suppose θ is a formula in the vocabulary L ∪ {P}, where P is an n-ary predicate symbol. Let Sb(θ, P, φ(x1 , . . . , xn )) be obtained from θ by replacing Pt1 . . . tn everywhere by φ(t1 , . . . , tn ). Lemma 3.25 (Preservation of equivalence under substitution) Suppose φ0 (x1 , . . . , xn ) and φ1 (x1 , . . . , xn ) are L-formulas of dependence logic such that φ0 (x1 , . . . , xn ) ≡∗ φ1 (x1 , . . . , xn ). Suppose θ is a formula in the vocabulary L ∪ {P}, where P is an n-ary predicate symbol. Then Sb(θ, P, φ0 (x1 , . . . , xn )) ≡∗ Sb(θ, P, φ1 (x1 , . . . , xn )). Proof The proof is straightforward. We use induction on θ . Let us use Sbd (θ) as a shorthand for Sb(θ, P, φd ). Atomic case. Suppose θ is Rt1 . . . tn . Now Sbd (θ) = φd . The claim follows from φ0 ≡∗ φ1 . Disjunction. Note that Sbd (φ ∨ ψ) = Sbd (φ) ∨ Sbd (ψ). Now (Sbd (φ ∨ ψ), X, 1) ∈ T if and only if (Sbd (φ) ∨ Sbd (ψ), X, 1) ∈ T if and only 2

In compositional semantics, roughly speaking, the meaning of a compound formula is completely determined by the way the formula is built from parts and by the meanings of the parts.

3.3 Logical equivalence and duality

33

if (X = Y ∪ Z such that (Sbd (φ), Y, 1) ∈ T and (Sbd (ψ), Z , 1) ∈ T ). By the induction hypothesis, this is equivalent to (X = Y ∪ Z such that (Sb1−d (φ), Y, 1) ∈ T and (Sb1−d (ψ), Z , 1) ∈ T ), i.e. (Sb1−d (φ) ∨ Sb1−d (ψ), X, 1) ∈ T , which is finally equivalent to (Sb1−d (φ ∨ ψ), X, 1) ∈ T . On the other hand, (Sbd (φ ∨ ψ), X, 0) ∈ T if and only if (Sbd (φ) ∨ Sbd (ψ), X, 0) ∈ T if and only if ((Sbd (φ), X, 0) ∈ T and (Sbd (ψ), X, 0) ∈ T ). By the induction hypothesis, this is equivalent to (Sb1−d (φ), X, 0) ∈ T and (Sb1−d (ψ), X, 0) ∈ T , i.e. (Sb1−d (φ) ∨ Sb1−d (ψ), X, 0) ∈ T , which is finally equivalent to (Sb1−d (φ ∨ ψ), X, 0) ∈ T . Negation. Sbe (¬φ) = ¬ Sbe (φ). Now (Sbe (¬φ), X, d) ∈ T if and only if (¬ Sbe (φ), X, d) ∈ T , which is equivalent to (Sbe (φ), X, 1 − d) ∈ T . By the induction hypothesis, this is equivalent to (Sb1−e (φ), X, 1 − d) ∈ T , i.e. (¬ Sb1−e (φ), X, d) ∈ T , and finally this is equivalent to (Sb1−e (¬φ), X, d) ∈ T . Existential quantification. Note that Sbd (∃xn φ) = ∃xn Sbd (φ). We may infer, as above, that (Sbd (∃xn φ), X, 1) ∈ T if and only if (∃xn Sbd (φ), X, 1) ∈ T if and only if there is F : X → M such that (Sbd (φ), X (F/xn ), 1) ∈ T . By the induction hypothesis, this is equivalent to the following: there is F : X → M such that (Sb1−d (φ), X (F/xn ), 1) ∈ T , i.e. to (∃xn Sb1−d (φ), X, 1) ∈ T , which is finally equivalent to (Sb1−d (∃xn φ), X, 1) ∈ T . On the other hand, (Sbd (∃xn φ), X, 0) ∈ T if and only if (∃xn Sbd (φ), X, 0) ∈ T , if and only if (Sbd (φ), X (M/xn ), 0) ∈ T . By the induction hypothesis, this is equivalent to (Sb1−d (φ), X (M/xn ), 0) ∈ T , i.e. (∃xn Sb1−d (φ), X, 0) ∈ T , which is finally equivalent to (Sb1−d (∃xn φ), X, 0) ∈ T .  As we did for first order logic in Chapter 2, we define two operations φ → φ p and φ → φ d by simultaneous induction, using the shorthands φ ∧ ψ and ∀xn φ: φ d = ¬φ if φ atomic, φ p = φ if φ atomic, (¬φ)d = φ p , (¬φ)p = φ d , (φ ∧ ψ)d = φ d ∨ ψ d , (φ ∧ ψ)p = φ p ∧ ψ p , (∃xn φ)d = ∀xn φ d , (∃xn φ)p = ∃xn φ p .

34

Dependence logic

We again call φ d the dual of φ, and φ p the negation normal form of φ. The basic results about duality in dependence logic, as in first order logic, are as given in Lemma 3.26. Lemma 3.26 φ ≡∗ φ p and ¬φ ≡∗ φ d . Thus the φ p operation is a mechanical method for translating a formula to a strongly logically equivalent formula in negation normal form. The dual operation is a mechanical method for translating a formula φ to one which is strongly logically equivalent to the negation of φ. We will see later (Section 7.3) that there is no hope of explaining φ ⇒ ψ in terms of a few simple rules. There are examples of φ and ψ such that to decide whether φ ⇒ ψ or not, one has to decide whether there are measurable cardinals in the set theoretic universe. Likewise, there are examples of φ and ψ such that to decide whether φ ⇒ ψ, one has to decide whether the Continuum Hypothesis holds. We examine next some elementary logical properties of formulas of dependence logic. Lemma 3.27 shows that the truth of a formula depends only on the interpretations of the variables occurring free in the formula. To this end, we define X V to be {sV : s ∈ X }. Lemma 3.27 Suppose V ⊇ Fr(φ). Then M |= X φ if and only if M |= X V φ. Proof Key to this result is the fact that t M s = t M sV  whenever Fr(t) ⊆ V . We use induction on φ to prove (φ, X, d) ∈ T ⇐⇒ (φ, X V, d) ∈ T whenever Fr(φ) ⊆ V . If φ is atomic, the claim is obvious, even in the case of =(t1 , . . . , tn ). Disjunction. Suppose (φ ∨ ψ, X, 1) ∈ T . Then X = Y ∪ Z such that (φ, Y, 1) ∈ T and (ψ, Z , 1) ∈ T . By the induction hypothesis, (φ, Y V, 1) ∈ T and (ψ, Z V, 1) ∈ T . Of course, X  V = (Y  V ) ∪ (Z  V ). Thus (φ ∨ ψ, X V, 1) ∈ T . Conversely, suppose (φ ∨ ψ, X V, 1) ∈ T . Then X  V = Y ∪ Z such that (φ, Y, 1) ∈ T and (φ, Z , 1) ∈ T . Choose Y  and Z  such that Y  V = Y , Z  V = Z and X = Y  ∪ Z  . Now we have (φ, Y  , 1) ∈ T and (ψ, Z  , 1) ∈ T by the induction hypothesis, Thus (φ ∨ ψ, X, 1) ∈ T . Suppose then (φ ∨ ψ, X, 0) ∈ T . Then (φ, X, 0) ∈ T and (ψ, X, 0) ∈ T . By the induction hypothesis (φ, X V, 0) ∈ T and (ψ, X V, 0) ∈ T . Thus (φ ∨ ψ, X V, 0) ∈ T . Conversely, suppose (φ ∨ ψ, X V, 0) ∈ T . Then (φ, X V, 0) ∈ T and (ψ, X V, 0) ∈ T . Now (φ, X, 0) ∈ T and (ψ, X, 0) ∈ T by the induction hypothesis. Thus (φ ∨ ψ, X, 0) ∈ T .

3.3 Logical equivalence and duality

35

Negation. Suppose (¬φ, X, d) ∈ T . Then (φ, X, 1 − d) ∈ T . By the induction hypothesis, (φ, X V, 1 − d) ∈ T . Thus (¬φ, X V, d) ∈ T . Conversely, suppose (¬φ, X V, d) ∈ T . Then (φ, X V, 1 − d) ∈ T . Now we have (φ, X, 1 − d) ∈ T by the induction hypothesis. Thus (¬φ, X, d) ∈ T . Existential quantification. Suppose (∃xn , X, 1) ∈ T . Then there is F : X → M such that (φ, X (F/xn ), 1) ∈ T . By the induction hypothesis, (φ, X (F/xn )W, 1) ∈ T , where W = V ∪ {n}. Note that X (F/xn )W = (X V )(F/xn ). Thus (∃xn φ, X V, 1) ∈ T . Conversely, suppose (∃xn φ, X V, 1) ∈ T . Then there is F : X V → M such that (φ, (X V )(F/xn ), 1) ∈ T . Again, X (F/xn )W = (X V )(F/xn ), and thus, by the induction hypothesis, (φ, X (F/xn ), 1) ∈ T , i.e. (∃xn φ, X, 1) ∈ T .  In Lemma 3.28 we have the restriction, familiar from substitution rules of first order logic, that in substitution no free occurrence of a variable should become a bound. Lemma 3.28 (Change of free variables) Let the free variables of φ be x1 , . . . , xn . Let i 1 , . . . , i n be distinct. Let φ  be obtained from φ by replacing x j everywhere by xi j , where j = 1, . . . , n. If X is an assignment set with dom(X ) = {1, . . . , n}, let X  consist of the assignments xi j → s(x j ), where s ∈ X . Then M |= X φ ⇐⇒ M |= X  φ  . Finally, we note the important fact that types are preserved by isomorphisms (recall the definition of M ∼ = M in Section 2.2). Lemma 3.29 (Isomorphism preserves truth) Suppose M ∼ = M . If φ ∈ D,  then M |= φ ⇐⇒ M |= φ. Proof Let T be the fundamental predicate (see Definition 3.5) for M and let T  be the fundamental predicate for M . Let π : M ∼ = M . For any assignment s for M, let πs be the assignment πs(xn ) = π (s(xn )). For any team X for M, let π X be the set of πs, where s ∈ X . We prove by induction on φ that for all teams X for M with the free variables of φ in its domain, and all d ∈ {0, 1}: (φ, X, d) ∈ T ⇐⇒ (φ, π X, d) ∈ T  . The claim is trivial for first order atomic formulas, and is clearly preserved by negation.

36

Dependence logic

Case (A) φ is =(t1 , . . . , tn ). Suppose (φ, X, 1) ∈ T , i.e. for all s, s  ∈ X M M such that t1M s = t1M s  , . . . , tn−1

s = tn−1

s  , we have  M M   tn s = tn s . Let π s, π s ∈ π X such that t1M π s =     M M t1M π s  , . . . , tn−1

πs = tn−1

πs  . As tiM π s = π tiM s M M  M M for all i, we have t1 s = t1 s , . . . , tn−1 s = tn−1

s  , hence   we also have tnM s = tnM s  , and finally tnM π s = tnM π s  , as desired. The converse is similar. Suppose then (φ, X, 0) ∈ T . This is equivalent to X = ∅, which is equivalent to π X = ∅, i.e. (φ, π X, 0) ∈ T  . Case (B) φ is ψ ∨ θ. Suppose (φ, X, 0) ∈ T . Then (ψ, X, 0) ∈ T and (θ, X, 0) ∈ T , whence (ψ, π X, 0) ∈ T  and (θ, π X, 0) ∈ T  , and finally (φ, π X, 0) ∈ T  . The converse is similar. Suppose then (φ, X, 1) ∈ T . Then X = X 0 ∪ X 1 such that (ψ, X 0 , 1) ∈ T and (θ, X 1 , 1) ∈ T . By the induction hypothesis, (ψ, π X 0 , 1) ∈ T  and (θ, π X 1 , 1) ∈ T  . Of course, π X = π X 0 ∪ π X 1 . Thus (φ, π X, 1) ∈ T  . Conversely, suppose (φ, π X, 1) ∈ T  . Then π X = X 0 ∪ X 1 such that (ψ, π X 0 , 1) ∈ T  and (θ, π X 1 , 1) ∈ T  . Let Y0 = {s ∈ X : π s ∈ X 0 } and Y1 = {s ∈ X : πs ∈ X 1 }. Thus π Y0 = X 0 and π Y1 = X 1 . By the induction hypothesis, M Y0 φ and M Y1 ψ. Since X = Y0 ∪ Y1 , we have (φ, X, 1) ∈ T . Case (C) φ is ∃xn ψ. This is left as an exercise.  Exercise 3.21 Prove the following non-equivalences: (i) (a) φ ∨ ¬φ ≡∗ , (b) φ ∧ ¬φ ≡∗ ¬ , but φ ∧ ¬φ ≡ ¬ ; (ii) (φ ∧ φ) ≡∗ φ, but (φ ∧ φ) ≡ φ; (iii) (φ ∨ φ) ≡∗ φ; (iv) (φ ∨ ψ) ∧ θ ≡∗ (φ ∧ θ ) ∨ (ψ ∧ θ); (v) (φ ∧ ψ) ∨ θ ≡∗ (φ ∨ θ ) ∧ (ψ ∨ θ). Note that each of these non-equivalences is actually an equivalence in first order logic. Exercise 3.22 Prove the following results familiar also from first order logic: (i) (a) ∃xn (φ ∧ ψ) ≡ (∃xn φ ∧ ∃xn ψ), but ∃xn (φ ∧ ψ) ⇒∗ (∃xn φ ∧ ∃xn ψ), (b) ∀xn (φ ∨ ψ) ≡ (∀xn φ ∨ ∀xn ψ), but (∀xn φ ∨ ∀xn ψ) ⇒∗ ∀xn (φ ∨ ψ); (ii) ∃xn ∀xm φ ≡∗ ∀xm ∃.xn φ. but ∃xn ∀xm φ ⇒∗ ∀xm ∃xn φ. Exercise 3.23 Prove Lemma 3.22. Exercise 3.24 Prove Lemma 3.23.

3.4 First order formulas

37

Exercise 3.25 Prove Lemma 3.26. Exercise 3.26 Prove Lemma 3.28. Exercise 3.27 Prove Case C in Lemma 3.29.

3.4 First order formulas Some formulas of dependence logic can be immediately recognized as first order merely by their appearance. They simply do not have any occurrences of the dependence formulas =(t1 , . . . , tn ) as subformulas. We then appropriately call them first order. Other formulas may be apparently non-first order, but turn out to be logically equivalent to a first order formula. Our goal in this section is to show that for apparently first order formulas our dependence logic truth definition (Definition 3.5 with X = ∅) coincides with the standard first order truth definition (Section 2.4). We also give a simple criterion called the Flatness Test that can be used to test whether a formula of dependence logic is logically equivalent to a first order formula. We begin by proving that a team is of a first order type φ if every assignment s in X satisfies φ. Note the a priori difference between an assignment s satisfying a first order formula φ and the team {s} being of type φ. We will show that these conditions are equivalent, but this indeed needs a proof. Proposition 3.30 If an L-formula φ of dependence logic is first order, then: (i) if M |=s φ for all s ∈ X , then (φ, X, 1) ∈ T ; (ii) if M |=s ¬φ for all s ∈ X , then (φ, X, 0) ∈ T . Proof We use induction as follows. If t1M s = t2M s for all s ∈ X , then (t1 = t2 , X, 1) ∈ T by (D1). If t1M s = t2M s for all s ∈ X , then (t1 = t2 , X, 0) ∈ T by (D2). (=(), X, 1) ∈ T by (D3). (=(), ∅, 0) ∈ T by (D4). If (t1M s, . . . , tnM s) ∈ R M for all s ∈ X , then (Rt1 . . . tn , X, 1) ∈ T by (D5). (6) If (t1M s, . . . , tnM s) ∈ / R M for all s ∈ X , then (Rt1 . . . tn , X, 0) ∈ T by (D6). (7) If M |=s ¬(φ ∨ ψ) for all s ∈ X , then M |=s ¬φ for all s ∈ X and M |=s ¬ψ for all s ∈ X , whence (φ, X, 0) ∈ T and (ψ, X, 0) ∈ T , and finally ((φ ∨ ψ), X, 0) ∈ T by (D7).

(1) (2) (3) (4) (5)

38

Dependence logic

(8) If M |=s φ ∨ ψ for all s ∈ X , then X = Y ∪ Z such that M |= φ for all s ∈ Y and M |= ψ for all s ∈ Z . Thus (ψ, Y, 1) ∈ T and (ψ, Z , 1) ∈ T , whence ((φ ∨ ψ), Y ∪ Z , 1) ∈ T by (D8). (9) If M |=s ¬φ for all s ∈ X , then (φ, X, 0) ∈ T , whence (¬φ, X, 1) ∈ T by (D9). (10) If M |=s ¬¬φ for all s ∈ x, then (φ, X, 1) ∈ T , whence (¬φ, X, 0) ∈ T by (D10). (11) If M |=s ∃xn φ for all s ∈ X , then for all s ∈ X there is as ∈ M such that M |=s(as /xn ) φ. Now (φ, {s}(F/xn ), 1) ∈ T for F : X → M such that F(s) = as . Thus (∃xn φ, X, 1) ∈ T . (12) If M |=s ¬∃xn φ for all s ∈ X , then for all a ∈ M we have for all s ∈ X M |=s(a/xn ) ¬φ. Now (φ, X (M/xn ), 0) ∈ T . Thus (∃xn φ, X, 0) ∈ T .  Now for the other direction. Proposition 3.31 If an L-formula φ of dependence logic is first order, then: (i) if (φ, X, 1) ∈ T , then M |=s φ for all s ∈ X ; (ii) if (φ, X, 0) ∈ T , then M |=s ¬φ for all s ∈ X . Proof We use induction as follows. If (t1 = t2 , X, 1) ∈ T , then t1M s = t2M s for all s ∈ X by (E1). If (t1 = t2 , X, 0) ∈ T , then t1M s = t2M s for all s ∈ X by (E2). (=(), X, 1) ∈ T and likewise M |=s for all s ∈ X . (=(), ∅, 0) ∈ T and likewise M |=s ¬ for all (i.e. none) s ∈ ∅. If (Rt1 . . . tn , X, 1) ∈ T , then (t1M s, . . . , tnM s) ∈ R M for all s ∈ X by (E5). (6) If (Rt1 . . . tn , X, 0) ∈ T , then (t1M s, . . . , tnM s) ∈ / R M for all s ∈ X by (E6). (7) If (φ ∨ ψ, X, 0) ∈ T , then (φ, X, 0) ∈ T and (ψ, X, 0) ∈ T by E7, whence M |=s ¬φ for all s ∈ X and M |=s ¬ψ for all s ∈ X , whence finally M |=s ¬(φ ∨ ψ) for all s ∈ X . (8) If (φ ∨ ψ, X, 1) ∈ T , then X = Y ∪ Z such that (φ, Y, 1) ∈ T and (ψ, Z , 1) ∈ T by (E8), whence M |=s φ for all s ∈ Y and M |=s ψ for all s ∈ Z , and therefore M |=s φ ∧ ψ for all s ∈ X . (1) (2) (3) (4) (5)

We leave the other cases as an exercise.



We are now ready to combine Propositions 3.30 and 3.31 in order to prove that the semantics we gave in Definition 3.5 coincide in the case of first order formulas with the more traditional semantics given in Section 2.4.

3.4 First order formulas

39

Corollary 3.32 Let φ be a first order L-formula of dependence logic. Then: (i) M |={s} φ if and only if M |=s φ; (ii) M |= X φ if and only if M |=s φ for all s ∈ X . Proof If M |={s} φ, then M |=s φ by Proposition 3.31. If M |=s φ, then M |={s} φ by Proposition 3.30.  We shall now introduce a test, comparable to the Closure Test introduced above. The Closure Test was used to test which types of teams are definable in dependence logic. With our new test we can check whether a type is first order, at least up to logical equivalence. Definition 3.33 (Flatness3 Test) We say that φ passes the Flatness Test if, for all M and X , M |= X φ ⇐⇒ (M |={s} φ for all s ∈ X ). Proposition 3.34 Passing the Flatness Test is preserved by logical equivalence. Proof Suppose φ ≡ ψ and φ passes the Flatness Test. Suppose M |={s} ψ for all s ∈ X . By logical equivalence, M |={s} φ for all s ∈ X . But φ passes the Flatness Test. So M |= X φ, and therefore, by our assumption, M |= X ψ.  Proposition 3.35 Any L-formula φ of dependence logic that is logically equivalent to a first order formula satisfies the Flatness Test. Proof Suppose φ ≡ ψ, where ψ is first order. Since ψ satisfies the Flatness Test, φ does also, by Proposition 3.34.  Example 3.36 =(x0 , x1 ) does not pass the Flatness Test, as the team X = {{(0, 0), (1, 1)}, {(0, 1), (1, 1)}} in a model M with at least two elements 0 and 1 shows. Namely, M |= X =(x0 , x1 ), but M |={s} =(x0 , x1 ) for s ∈ X . We conclude that =(x0 , x1 ) is not logically equivalent to a first order formula. Example 3.37 ∃x2 (=(x0 , x2 ) ∧ x2 = x1 ) does not pass the Flatness Test, as the team X = {s, s  }, s = {(0, 0), (1, 1)}, s  = {(0, 0), (1, 0)} in a model M with at least two elements 0 and 1 shows. Namely, if F : X → M witnesses M |= X (F/x2 ) =(x0 , x2 ) ∧ x2 = x1 , then s(x0 ) = s  (x0 ), but 1 = s(x1 ) = F(s) = F(s  ) = s  (x1 ) = 0,

3

Hodges defines a flattening operation in ref. [21], hence the word ‘Flatness’.

Dependence logic

40

a contradiction. We conclude that ∃x2 (=(x0 , x2 ) ∧ x2 = x1 ) is not logically equivalent to a first order formula. Example 3.38 Let L = {+, ·, 0, 1, <} and M = (N, +, ·, 0, 1, <), the standard model of arithmetic. The formula ∃x0 (=(x0 ) ∧ (x1 < x0 )) fails to meet the Flatness Test. To see this, we first note that if X = {s}, then X is of the type of the formula, as we can choose as to be equal to s(x1 ) + 1. On the other hand, let X = {sn : n ∈ N}, where sn (x1 ) = n. It is impossible to choose a such that a > sn (x1 ) for all n ∈ N. Exercise 3.28 Finish the proof of Proposition 3.31. Exercise 3.29 Find in each case a logically equivalent first order formula: (i) (ii) (iii) (iv) (v)

∃x0 (=(x1 , x0 ) ∧ P x0 ), ∃x0 (=(x1 , x0 ) ∧ P x1 ), ∃x0 ((=(x2 , x0 ) ∧ P x0 ) → P x1 ), ∃x0 (=(x1 , x0 ) ∧ Rx0 x1 ), ∃x0 (=(x0 ) ∧ (Rx1 x2 ∨ Rx0 x0 )).

Exercise 3.30 Find in each case a logically equivalent first order formula: (i) ∃x0 (=(x1 , x0 ) ∧ ( f x1 = x1 )), (ii) ∃x0 (=(x2 , x0 ) ∧ (P f x0 ∧ ¬P x1 )). Exercise 3.31 Which of the following formulas are logically equivalent to a first order formula? (i) =() ∨ ¬ =(); (ii) =(x0 ); (iii) =(x0 , x0 ). Exercise 3.32 Which of the following formulas are logically equivalent to a first order formula? (i) =(x0 , x1 , x2 ) ∧ x0 = x1 ; (ii) (=(x0 , x2 ) ∧ x0 = x1 ) → =(x1 , x2 ); (iii) =(x0 , x1 , x2 ) ∨ ¬ =(x0 , x1 , x2 ). Exercise 3.33 Which of the following formulas are logically equivalent to a first order formula? (i) ∀x0 ∃x2 (=(x0 , x2 ) ∧ x2 = x1 ); (ii) ∀x1 ∃x2 (=(x0 , x2 ) ∧ x2 = x1 ); (iii) ∀x0 ∀x1 ∃x2 (=(x0 , x2 ) ∧ x2 = x1 ).

3.4 First order formulas

41

Exercise 3.34 Which of the following formulas are logically equivalent to a first order formula? (i) (ii) (iii) (iv)

∀x0 (x0 = x1 → =(x0 , x1 )); ∀x0 ( f x0 = x1 → =(x0 , x1 )); ∀x0 (x0 = f x1 → =(x0 , x1 )); ∀x1 ∀x0 ( f x0 = f x1 → =(x0 , x1 )).

Exercise 3.35 Let L = ∅ and let M be an L-structure with M = {0, 1}. Show that the following types of a team X with domain {x0 , x1 , x2 } are non-first order: (i) ∃x0 (=(x2 , x0 ) ∧ ¬(x0 = x1 )), (ii) ∃x0 (=(x2 , x0 ) ∧ (x0 = x1 ∨ x0 = x2 )), (iii) ∃x0 (=(x2 , x0 ) ∧ (x0 = x1 ∧ ¬x0 = x2 )). Exercise 3.36 Let L = {<} and let M be an L-structure (N, <). Show that the following types of a team X with domain {x0 , x1 , x2 } are non-first order: (i) ∃x0 (=(x2 , x0 ) ∧ ¬(x0 = x1 )), (ii) ∃x0 (x0 < x1 ∧ x0 < x2 ∧ =(x2 , x0 )), (iii) ∃x0 (x0 < x2 ∧ ¬(x0 = x1 ) ∧ =(x2 , x0 )). Exercise 3.37 Let L = {R}, #(R) = 1. Find an L-structure M which demonstrates that the following properties of a team X with domain {x0 , x1 , x2 } are non-first order: (i) ∃x0 (Rx0 ∧ =(x1 , x0 ) ∧ ¬x0 = x2 ), (ii) ∃x0 (=(x2 , x0 ) ∧ (Rx0 ↔ Rx1 )), (iii) ∃x0 (=(x2 , x0 ) ∧ ((Rx1 ∧ ¬Rx0 ) ∨ (¬Rx1 ∧ Rx0 ))). Exercise 3.38 Let L = {R}, #(R) = 2. Find an L-structure M which demonstrates that the following properties of a team X with domain {x0 , x1 , x2 } are non-first order: (i) ∃x0 (=(x2 , x0 ) ∧ Rx0 x1 ), (ii) ∃x0 (=(x2 , x0 ) ∧ (Rx1 x0 ∧ Rx2 x0 )), (iii) ∃x0 (=(x2 , x0 ) ∧ ((Rx1 x0 ) ↔ (Rx2 x0 ))). Exercise 3.39 Let L = { f }, #( f ) = 1. Find an L-structure M which demonstrates that the following properties of a team X with domain {x1 } are non-first order: (i) ∃x0 (=(x0 ) ∧ ( f x1 = x0 )), (ii) ∃x0 ∃x0 (=(x0 ) ∧ ( f x0 = x1 )), (iii) ∃x0 ∃x0 (=(x0 ) ∧ ( f x0 = f x1 )).

42

Dependence logic

Exercise 3.40 A formula φ of dependence logic is coherent if the following holds: Any team X is of type φ if and only if for every s, s  ∈ X the pair team {s, s  } is of type φ. Note that the formula (=(x1 , . . . , xn ) ∧ φ) is coherent if φ is. Show that for every first order φ with Fr(φ) = {x1 }, the type ∃x0 (=(x1 , x0 ) ∧ φ) is coherent. Exercise 3.41 Give an example of a formula φ of dependence logic which is not coherent (see Exercise 3.40 for the definition of coherence).

3.5 The flattening technique We now introduce a technique which may seem frivolous at first sight but proves very useful in the end. This is the process of flattening, by which we mean getting rid of the dependence formulas =(t1 , . . . , tn ). Naturally we lose something, but this method reveals whether a formula has genuine occurrences of dependence or just ersatz ones. Definition 3.39 The flattening φ f of a formula φ of dependence logic is defined by induction as follows: (t1 = t2 )f = (t1 = t2 ), (Rt1 . . . tn )f = Rt1 . . . tn , (=(t1 , . . . , tn ))f = , (¬φ)f = ¬φ f , (φ ∨ ψ)f = φ f ∨ ψ f , (∃xn φ)f = ∃xn φ f . Note that the result of flattening is always first order. The main feature of flattening is that it preserves truth. Proposition 3.40 If φ is an L-formula of dependence logic, then φ ⇒ φ f . Proof Inspection of Definition 3.5 reveals immediately that in each case where (φ, X, d) ∈ T , we also have (φ f , X, d) ∈ T .  We can use Proposition 3.40 to prove various useful little results which are often comforting in enforcing our intuition. We first point out that although a team may be of the type of both a formula and its negation, this can only happen if the team is empty and thereby is of the type of any formula. Corollary 3.41 M |= X (φ ∧ ¬φ) if and only if X = ∅.

3.5 The flattening technique

43

Proof We already know that M |=∅ (φ ∧ ¬φ). On the other hand, if M |= X (φ ∧ ¬φ) and s ∈ X , then M |=s (φ f ∧ ¬φ f ), a contradiction.  Corollary 3.42 (Modus Ponens) Suppose M |= X φ → ψ and M |= X φ. Then M |= X ψ. Proof M |= X ¬φ ∨ ψ implies X = Y ∪ Z such that M |=Y ¬φ and M |= Z ψ. Now M |=Y φ and M |=Y ¬φ, whence Y = ∅. Thus X = Z and M |= X ψ follows.  In general, we may conclude from Proposition 3.40 that a non-empty team cannot have the type of a formula which is contradictory in first order logic when flattened. When all the subtle properties of dependence logic are laid bare in front of us, we tend to seek solace in anything solid, anything that we know for certain from our experience in first order logic. Flattening is one solace. By simply ignoring the dependence statements =(t1 , . . . , tn ), we can recover in a sense the first order content of the formula. When we master this technique, we begin to understand the effect of the presence of dependence statements in a formula. Example 3.43 No non-empty team can have the type of any of the following formulas, whatever formulas of dependence logic the formulas φ and ψ are: φ(c) ∧ ∀x0 ¬φ(x0 ), ∀x0 ¬φ ∧ ∀x0 ¬ψ ∧ ∃x0 (φ ∨ ψ), ¬(((φ → ψ) → φ) → φ). The flattenings of these formulas are, respectively, given by φ f (c) ∧ ∀x0 ¬φ f (x0 ), ∀x0 ¬φ f ∧ ∀x0 ¬ψ f ∧ ∃x0 (φ f ∨ ψ f ), ¬(((φ f → ψ f ) → φ f ) → φ f ), none of which can be satisfied by any assignment in first order logic. In the last case one can use truth tables to verify this. As the previous example shows, the Truth Table Method, so useful in propositional calculus, also has a role in dependence logic. Exercise 3.42 Let φ be the formula ∃x0 ∀x1 ¬(=(x2 , x1 ) ∧ (x0 = x1 )) of D. Show that the flattening of φ is not a strong logical consequence of φ. Exercise 3.43 Show that if M |= X (φ → ψ) and M |= X ¬ψ, then M |= X ¬φ.

44

Dependence logic

Exercise 3.44 Show that no non-empty team can have the type of any of the following formulas: (i) (ii) (iii) (iv)

¬ =(x0 , x1 ), ¬(=(x0 , x1 ) → =(x2 , x1 )), ¬ =( f x0 , x0 ) ∨ ¬ =(x0 , f x0 ), ∀x0 ∃x1 ∀x2 ∃x3 ¬(φ → =(x0 , x1 )).

Exercise 3.45 Explain the difference between teams of type =(x0 , x2 ) ∧ =(x1 , x2 ) and teams of type =(x0 , x1 , x2 ). Exercise 3.46 Show that if |= φ → ψ, then φ ⇒∗ ψ. Exercise 3.47 Show that if φ has only x0 and x1 free, then ∀x0 ∃x1 φ ⇒ ∀x0 ∃x1 (=(x0 , x1 ) ∧ φ). Exercise 3.48 Show that the formulas ∀x0 ∃x1 ∀x2 ∃x3 (=(x0 , x1 ) ∧ =(x2 , x3 ) ∧ φ) and ∀x2 ∃x3 ∀x0 ∃x1 (=(x0 , x1 ) ∧ =(x2 , x3 ) ∧ φ), where φ is first order, are logically equivalent. Exercise 3.49 Prove the following: (i) ∃xn (φ ∧ ψ) ≡∗ φ ∧ ∃xn ψ if xn not free in φ; (ii) ∀xn (φ ∨ ψ) ≡∗ φ ∨ ∀xn ψ if xn not free in φ. Exercise 3.50 Prove that |= ∃x1 (=(x1 ) ∧ x1 = c) but |= ∀x0 ∃x1 (=(x1 ) ∧ x1 = x0 ). Exercise 3.51 (Prenex normal form) A formula of dependence logic is in prenex normal form if all quantifiers are in the beginning of the formula. Use Lemma 3.23 and Exercise 3.49 to prove that every formula of dependence logic is strongly equivalent to a formula which has the same free variables and is in the prenex normal form.

3.6 Dependence/independence friendly logic We review the relation of our dependence logic D to the independence friendly logics of refs. [19]–[21]. The backslashed quantifier, ∃xn \{xi0 , . . . , xim−1 }φ,

(3.4)

3.6 Dependence/independence friendly logic

45

introduced in ref. [20], with the intuitive meaning “there exists xn , depending only on xi0 . . . xim−1 , such that φ,”

(3.5)

can be defined in dependence logic by the following formula: ∃xn (=(xi0 , . . . , xim−1 , xn ) ∧ φ).

(3.6)

Conversely, we can define =(xi0 , . . . , xim−1 , xm ) in terms of Eq. (3.4) by means of the following formula: ∃xn \{xi0 , . . . , xim−1 }(xn = xm ).

(3.7)

Similarly, we can define =(t1 , . . . , tn ) in terms of Eq. (3.4), when t1 , . . . , tn are terms. Dependence friendly logic, denoted DF, is the fragment of dependence logic obtained by leaving out the atomic dependence formulas =(t1 , . . . , tn ) and adding all the backslashed quantifiers (Eq. (3.4)). Dependence logic and DF have the same expressive power, not just on the level of sentences, but even on the level of formulas in the following sense. Proposition 3.44 (i) For every φ in D, there is φ ∗ in DF so that, for all models M and all teams X, M |= X φ ⇐⇒ M |= X φ ∗ . (ii) For every ψ in DF, there is ψ ∗∗ in D so that, for all models M and all teams X , M |= X ψ ⇐⇒ M |= X ψ ∗∗ . We can base the study of dependence either on the atomic formulas t1 = tn , Rt1 . . . tn , =(t1 , . . . , tn ), together with the logical operations ¬, ∨, ∃xn , as we have done in this book, or on the atomic formulas t1 = tn , Rt1 . . . tn , together with the logical operations ¬, ∨, ∃xn \{xi0 , . . . , xim−1 }. The results of this book remain true if D is replaced by DF. The slashed quantifier, ∃xn /{xi0 , . . . , xim−1 }φ,

(3.8)

used in ref. [21] has the following intuitive meaning: “there exists xn , independently of xi0 . . . xim−1 , such that φ,”

(3.9)

which we take to mean “there exists xn , depending only on variables other than xi0 . . . xim−1 , such that φ,”

(3.10)

Dependence logic

46

If the other variables, referred to in Eq. (3.10) are x j0 . . . x jl−1 , then Eq. (3.9) is intuitively equivalent to ∃xn \{x j0 , . . . , x jl−1 }φ.

(3.11)

Independence friendly logic, denoted IF, is the fragment of dependence logic obtained by leaving out the atomic dependence formulas =(t1 , . . . , tn ) and adding all the slashed quantifiers from Eq. (3.8) with Eq. (3.9) (or rather Eq. (3.11)) as their meaning. Sentences of dependence logic and IF have the same expressive power in the following sense: (i) For every sentence φ in D, there is a sentence φ ∗ in IF so that, for all models M, M |= φ ⇐⇒ M |= φ ∗ . (ii) For every sentence ψ in IF, there is a sentence ψ ∗∗ in D so that, for all models M, M |= ψ ⇐⇒ M |= ψ ∗∗ . We observed that we can base the study of dependence on D or DF and everything will go through more or less in the same way. However, IF differs more from D than DF, even if the expressive power is in the above sense the same as that of D, and even if there is the intuitive equivalence of Eqs. (3.9) and (3.11). Dealing with Eq. (3.10) rather than Eq. (3.5) involves the complication that one has to decide whether “other variable” refers to other variables actually appearing in a formula φ, or to other variables in the domain of the team X under consideration. In the latter case, variables not occurring in the formula φ may still determine whether the team X is of type φ. Consider, for example, the following formula: ∃x0 /{x1 }(x0 = x1 ).

(3.12)

The teams in Table 3.10 are of the type given in Eq. (3.12) as we can let x0 depend on x2 . The variable x2 , which does not occur in Eq. (3.12), signals Table 3.10. x0

x1

x2

x0

x1

x2

1 1 1

1 3 8

1 3 8

1 1 1

1 3 8

5 2 1

3.6 Dependence/independence friendly logic

47

Table 3.11. x0

x1

x2

1 1 1

1 3 8

5 5 5

what x1 is. However, the team in Table 3.11 is not of the type given in Eq. (3.12), even though all three teams agree on all variables that occur in Eq. (3.12). The corresponding formula, ∃x0 \{x2 }(x0 = x1 ),

(3.13)

of DF avoids this as all variables that are actually used are mentioned in the formula. In this respect, DF is easier to work with than IF. Exercise 3.52 Give a logically equivalent formula in D for the DF-formula ∃x2 \x1 Rx1 x2 . Exercise 3.53 Give for both of the following D-formulas: (i) ∃x2 ∃x3 (=(x0 , x2 ) ∧ =(x1 , x3 ) ∧ Rx0 x1 x2 x3 ), (ii) =(x0 ) ∨ =(x1 ), a logically equivalent formula in DF. Exercise 3.54 Give for each of the following D-sentences φ: (i) ∀x0 ∃x1 (¬ =(x0 , x1 ) ∧ ¬x1 = x0 ), (ii) ∀x0 ∀x1 ∃x2 (=(x1 , x2 ) ∧ ¬x2 = x1 ). a sentence φ ∗ in IF so that φ and φ ∗ have the same models. Exercise 3.55 Give for both of the following IF-sentences: (i) ∀x0 ∃x1 /{x0 }(x0 = x1 ), (ii) ∀x0 ∃x1 /{x0 }(x1 ≤ x0 ). a first order sentence with the same models. Exercise 3.56 Give a definition of =(t1 , . . . , tn ) in DF. Exercise 3.57 Prove Proposition 3.44.

4 Examples

We now study some more complicated examples involving many quantifiers. In all these examples we use quantifiers to express the existence of some functions. There is a certain easy trick for accomplishing this which hopefully becomes apparent to the reader. The main idea is that some variables stand for arguments and some stand for values of functions that the sentence stipulates to exist.

4.1 Even cardinality On a finite set {a1 , . . . , an } of even size, one can define a one-to-one function f which is its own inverse and has no fixed points, as in the following picture:

Conversely, any finite set with such a function has even cardinality. In the following sentence we think of f (x0 ) as x1 and of f (x2 ) as x3 . So x1 depends only on x0 , and x3 depends only on x2 , which is guaranteed by =(x2 , x3 ). To make sure f has no fixed points we stipulate ¬(x0 = x1 ). The condition (x1 = x2 → x3 = x0 ) says in effect ( f (x0 ) = x2 → f (x2 ) = x0 ), i.e. f ( f (x0 )) = x0 . Let even : ∀x0 ∃x1 ∀x2 ∃x3 (=(x2 , x3 ) ∧ ¬(x0 = x1 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x2 → x3 = x0 )). We claim that the sentence even of dependence logic is true in a finite structure 48

4.1 Even cardinality

49

Table 4.1. The teams X and X (F/x3 ), when M = {a1 , . . . , a4 }.

s1 s2 s3 s4 s5 s6 s7 s8 s9 s10 s11 s12 s13 s14 s15 s16

x0

x1

x2

a1 a1 a1 a1 a2 a2 a2 a2 a3 a3 a3 a3 a4 a4 a4 a4

a2 a2 a2 a2 a1 a1 a1 a1 a4 a4 a4 a4 a3 a3 a3 a3

a1 a2 a3 a4 a1 a2 a3 a4 a1 a2 a3 a4 a1 a2 a3 a4

s1 (F/x3 ) s2 (F/x3 ) s3 (F/x3 ) s4 (F/x3 ) s5 (F/x3 ) s6 (F/x3 ) s7 (F/x3 ) s8 (F/x3 ) s9 (F/x3 ) s10 (F/x3 ) s11 (F/x3 ) s12 (F/x3 ) s13 (F/x3 ) s14 (F/x3 ) s15 (F/x3 ) s16 (F/x3 )

x0

x1

x2

x3

a1 a1 a1 a1 a2 a2 a2 a2 a3 a3 a3 a3 a4 a4 a4 a4

a2 a2 a2 a2 a1 a1 a1 a1 a4 a4 a4 a4 a3 a3 a3 a3

a1 a2 a3 a4 a1 a2 a3 a4 a1 a2 a3 a4 a1 a2 a3 a4

a2 a1 a4 a3 a2 a1 a4 a3 a2 a1 a4 a3 a2 a1 a4 a3

if and only if the size of the structure is even. To this end, suppose first M is a finite structure with M = {a1 , . . . , a2n }. Let X = {{(x0 , ai ), (x1 , f (ai )), (x2 , a j )} : 1 ≤ i ≤ n, 1 ≤ j ≤ n}, where f (ai ) = ai+1 if i is odd and f (ai ) = ai−1 if i is even (see Table 4.1). Let F : X → M such that F({(x0 , ai ), (x1 , f (ai )), (x2 , a j )}) = f (a j ). Now we note that M |= X (F/x3 ) (=(x2 , x3 ) ∧ ¬(x0 = x1 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x2 → x3 = x0 )), and therefore M |= X ∃x3 (=(x2 , x3 ) ∧ ¬(x0 = x1 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x2 → x3 = x0 )).

Examples

50

Table 4.2. The teams Y and Y (G/x1 ) x0 s1 s5 s9 s13

a1 a2 a3 a4

s1 (G/x1 ) s5 (G/x1 ) s9 (G/x1 ) s13 (G/x1 )

x0

x1

a1 a2 a3 a4

a2 a1 a4 a3

Let Y = {{(x0 , ai )} : 1 ≤ i ≤ n} and let G : X → M G({(x0 , ai )}) = f (ai ).

be such that

M |=Y (G/x1 ) ∀x2 ∃x3 (=(x2 , x3 ) ∧ ¬(x0 = x1 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x2 → x3 = x0 )), whence M |=Y ∃x1 ∀x2 ∃x3 (=(x2 , x3 ) ∧ ¬(x0 = x1 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x2 → x3 = x0 )), and finally M |={∅} even . See Table 4.2. For the converse, suppose X = {∅} and M |= X even . Thus there are F : X (M/x0 ) → M and1 G : X (M/x0 , F/x1 , M/x2 ) → M such that M |= X (M/x0 ,F/x1 ,M/x2 ,G/x3 ) (=(x2 , x3 ) ∧ ¬(x0 = x1 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x2 → x3 = x0 )). Let F({(x0 , a)}) = f (a) for each a ∈ M. Then G({(x0 , a), (x1 , f (a)), (2, b)}) = f (b). It follows that f is one-to-one, f (a) = a and f ( f (a)) = a for all a. Thus, if M is finite, its cardinality must be even. Note that even cardinality is not expressible in first order logic, as a simple application of the Compactness Theorem2 (or Ehrenfeucht–Fra¨ıss´e games) shows. 1 2

We compose X (M/x0 )(F/x1 )(M/x2 ) into X (M/x0 , F/x1 , M/x2 ). Suppose a first order φ expresses even cardinality in finite models of the empty vocabulary. By the Compactness Theorem (see Section 6.2), both φ and ¬φ have an infinite model. By the Downward L¨owenheim–Skolem Theorem (see Section 6.2), φ has a countably infinite model M, and similarly ¬φ has a countably infinite model N . But since the vocabulary is empty, M∼ = N . This contradicts M |= φ and N |= ¬φ.

4.2 Cardinality

51

Exercise 4.1 Give a sentence of dependence logic which is true in a finite structure if and only if the size of the structure is odd. Note that ¬even would not do. Exercise 4.2 Give a sentence of dependence logic which is true in a finite structure if and only if the size of the structure is divisible by three.

4.2 Cardinality The domain of a structure is infinite if and only if there is a one-to-one function that maps the domain into a proper subset. For example, if the domain contains an infinite set A = {a0 , a1 , . . .}, we can map A onto the proper subset {a1 , a2 , . . .} with the mapping an → an+1 and the outside of A onto itself by the identity mapping. On the other hand, if f : M → M is a one-to-one function that does not have a in its range, then {a, f (a), f ( f (a)), . . .} is an infinite subset:

In the following sentence we think of f (x0 ) as x1 and of g(x2 ) as x3 . So x1 depends only on x0 , and x3 depends only on x2 , which is guaranteed by =(x2 , x3 ). The condition ¬(x1 = x4 ) says x4 is outside the range of the function f . To make sure that f = g, we stipulate (x0 = x2 → x1 = x3 ). The condition (x1 = x3 → x0 = x2 ) says f is one-to-one. Let ∞ : ∃x4 ∀x0 ∃x1 ∀x2 ∃x3 (=(x2 , x3 ) ∧ ¬(x1 = x4 ) ∧ (x0 = x2 ↔ x1 = x3 )). We conclude that ∞ is true in a structure if and only if the domain of the structure is infinite. Exercise 4.3 A graph is a pair M = (G, E), where G is a set of elements called vertices and E is an anti-reflexive symmetric binary relation on G called the edge-relation. The degree of a vertex is the number of vertices that are connected by a (single) edge to v. The degree of v is said to be infinite if the set of vertices that are connected by an edge to v is infinite. Give a sentence of dependence logic which is true in a graph if and only if every vertex has infinite degree.

52

Examples

Exercise 4.4 Give a sentence of dependence logic which is true in a graph if and only if the graph has infinitely many isolated vertices. (A vertex is isolated if it has no neighbors.) Exercise 4.5 Give a sentence of dependence logic which is true in a graph if and only if the graph has infinitely many vertices of infinite degree. (The degree of a vertex is the cardinality of the set of neighbors of the vertex.) A more general question about cardinality is equicardinality. In this case we have two unary predicates P and Q on a set M and we want to know whether they have the same cardinality; that is, whether there is a bijection f from P to Q. In the following sentence = we think of f (x0 ) as x1 and of f (x2 ) as x3 : = : ∀x0 ∃x1 ∀x2 ∃x3 (=(x2 , x3 ) ∧ ((P x0 ∧ Qx2 ) → (Qx1 ∧ P x3 ∧ (x0 = x3 ↔ x1 = x2 )))). Suppose we want to test whether a unary predicate Q has at least as many elements as another unary predicate P. Here we can use a simplification of = : ≤ : ∀x0 ∃x1 ∀x2 ∃x3 (=(x2 , x3 ) ∧ (P x0 → (Qx1 ∧ (x0 = x2 ↔ x1 = x3 )))). On the other hand, the following variant of = clearly expresses the isomorphism of two linear orders (P, < P ) and (Q, < Q ): ∼ = : ∀x 0 ∃x 1 ∀x 2 ∃x 3 (=(x 2 , x 3 ) ∧ ((P x 0 ∧ Qx 2 ) → ∧ (Qx1 ∧ P x3 ∧ ∧ (x0 < P x3 ↔ x1 < Q x2 )))). An isomorphism M → M is called an automorphism. The identity mapping is, of course, always an automorphism. An automorphism is non-trivial if it is not the identity mapping. Below is a picture of a finite structure with a non-trivial automorphism:

A structure is rigid if it has only one automorphism, namely the identity. Finite linear orders and, for example, (N, <) are rigid,3 but, for example, (Z, <) is 3

Any automorphism has to map the first element to the first element, the second element to the second element, the third element to the third element, etc.

4.3 Completeness

53

non-rigid. We can express the non-rigidity of a linear order with the following sentence: nr : ∃x4 ∀x0 ∃x1 ∀x2 ∃x3 (=(x2 , x3 ) ∧ (x0 = x4 → ¬(x1 = x4 )) ∧ (x0 < x3 ↔ x1 < x2 )) Exercise 4.6 Write down a sentence of D which is true in a group4 if and only if the group is non-rigid. Exercise 4.7 Write down a sentence of D which is true in a finite structure M if and only if the unary predicate P contains in M at least half of the elements of M. Exercise 4.8 A natural number n is a prime power if and only if there is a finite field of n elements. Use this fact to write down a sentence of D in the empty vocabulary which has finite models of exactly prime power cardinalities. Exercise 4.9 A group (G, ◦, e) is right orderable if there is a partial order ≤ in the set G such that x ≤ y implies x ◦ z ≤ y ◦ z for all x, y, z in G. Write down a sentence of D which is true in a group if and only if the group is right orderable. Exercise 4.10 An abelian group (G, +, 0) is the additive group of a field if there is a binary operation · on G and an element 1 in G such that (G, +, ·, 0, 1) is a field. Write down a sentence of D which is true in an abelian group if and only if the group is the additive group of a field.

4.3 Completeness Suppose we want to test whether a linear order < on a set M is complete or not, i.e. whether every non-empty A ⊆ M with an upper bound has a least upper bound. Since we have to talk about arbitrary subsets A of a domain M, we use a technique called guessing. This is simply fixing an element a of M and then taking an arbitrary function from M to M. We call a the “head” as if we were tossing a coin. The set A corresponds to the set of elements of M mapped to the head. For simplicity, we take the head to be an upper bound of A, which we assume to exist anyway. 4

A group is a structure (G, ◦, e) with a binary function ◦ and a constant e such that (1) for all a, b, c ∈ G: (a ◦ b) ◦ c = a ◦ (b ◦ c), (2) for all a ∈ G: e ◦ a = a ◦ e = a, (3) for all a ∈ G there is b ∈ G such that a ◦ b = b ◦ a = e. The group is abelian if, in addition, (4) for all a, b ∈ G: a ◦ b = b ◦ a.

54

Examples

A linear order is incomplete if and only if there is a non-empty initial segment A without a last point but with an upper bound such that for every element not in A there is a smaller element not in A. To express this we use the following sentence: cmpl : ∃x6 ∃x7 ∀x0 ∃x1 ∀x2 ∃x3 ∀x4 ∃x5 ∀x8 ∃x9 (=(x2 , x3 ) ∧ =(x4 , x5 ) ∧ =(x8 , x9 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x6 → x0 < x6 ) ∧ (x0 = x7 → x1 = x6 ) ∧ ((x0 < x2 ∧ x3 = x6 ) → x1 = x6 ) ∧ ((¬(x1 = x6 ) ∧ x0 = x4 ∧ x2 = x5 ) → (x5 < x0 ∧ ¬(x3 = x6 ))) ∧ ((x0 = x8 ∧ x1 = x6 ∧ x2 = x9 ) → (x8 < x9 ∧ x3 = x6 ))) The sentence cmpl is true in a linear order if and only if the linear order is incomplete. (cmpl is not necessarily the simplest one with this property.) Explanation of cmpl : The mapping x0 → x1 is the guessing function and x2 → x3 is a copy of it, as witnessed by (x0 = x2 → x1 = x3 ). x6 is the head, therefore we have (x1 = x6 → x0 < x6 ). x7 manifests non-emptiness of the guessed initial segment as witnessed by (x0 = x7 → x1 = x6 ). The clause ((x0 < x2 ∧ x3 = x6 ) → x1 = x6 ) guarantees the guessed set is really an initial segment. Finally we need to say that if an element x0 is above the initial segment (x1 = x6 ) then there is a smaller element x5 also above the initial segment. The mapping x8 → x9 makes sure the initial segment does not have a maximal element. The sentence cmpl has many quantifier alternations but that is not really essential as we could equivalently use the universal-existential sentence: cmpl : ∀x0 ∀x2 ∀x4 ∀x8 ∃x1 ∃x3 ∃x5 ∃x6 ∃x7 ∃x9 (=(x0 , x1 ) ∧ =(x2 , x3 ) ∧ =(x4 , x5 ) ∧ =(x6 ) ∧ =(x7 ) ∧ =(x8 , x9 )

4.4 Well-foundedness

55

∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x6 → x0 < x6 ) ∧ (x0 = x7 → x1 = x6 ) ∧ ((x0 < x2 ∧ x3 = x6 ) → x1 = x6 ) ∧ ((¬(x1 = x6 ) ∧ x0 = x4 ∧ x2 = x5 ) → (x5 < x0 ∧ ¬(x3 = x6 ))) ∧ ((x0 = x8 ∧ x1 = x6 ∧ x2 = x9 ) → (x8 < x9 ∧ x3 = x6 ))). Exercise 4.11 Give a sentence of D which is true in a linear order if and only if the linear order is isomorphic to a proper initial segment of itself.

4.4 Well-foundedness A binary relation R on a set M is well-founded if and only if there is no sequence a0 , a1 , . . . in M such that an+1 Ran for all n; otherwise it is ill-founded. An equivalent definition of well-foundedness is that there is no non-empty subset X of M such that for every element a in X there is an element b of X such that b Ra. To express ill-foundedness, we use the following sentence: wf : ∃x6 ∃x7 ∀x0 ∃x1 ∀x2 ∃x3 ∀x4 ∃x5 (=(x2 , x3 ) ∧ =(x4 , x5 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x0 = x7 → x1 = x6 ) ∧ ((x1 = x6 ∧ x0 = x4 ∧ x2 = x5 ) → (x3 = x6 ∧ Rx5 x4 ))). The sentence wf is true in a binary structure (M, R) if and only if R is illfounded. The explanation is as follows. The mapping x0 → x1 guesses the set X as the pre-image of x6 . The mapping x2 → x3 is a copy of the mapping x0 → x1 , as witnessed by (x0 = x2 → x1 = x3 ); x7 manifests non-emptiness of the guessed initial segment as witnessed by (x0 = x7 → x1 = x6 ). The clause ((x1 = x6 ∧ x0 = x4 ∧ x2 = x5 ) → (x3 = x6 ∧ Rx5 x4 )) guarantees the guessed set has no R-smallest element. Exercise 4.12 A partially ordered set is an L-structure M = (M, ≤M ) for the vocabulary L = {≤}, where ≤M is assumed to be reflexive (x ≤ x), transitive

56

Examples

(x ≤ y ≤ z ⇒ x ≤ z), and anti-symmetric (x ≤ y ≤ x ⇒ x = y). We shorten (x ≤M y & x = y) to x <M y. A chain of a partial order is a subset of M which is linearly ordered by ≤M . Give a sentence of D which is true in a partially ordered set if and only if the partial order has an infinite chain. Exercise 4.13 A tree is a partially ordered set M such that the set {x ∈ M : x <M t} of predecessors of any t ∈ M is well ordered by ≤M and there is a unique smallest element in M, called the root of the tree. Thus for any t <M s in M there is an immediate successor r of t such that t <M r ≤M s. A subtree of a tree is a substructure which is a tree. A tree is binary if every element has at most two immediate successors, and is a full binary tree if every element has exactly two immediate successors. Give a sentence of D which is true in a tree if and only if the tree has a full binary subtree. Exercise 4.14 The cofinality of a linear order is the smallest cardinal κ such that the order has an unbounded subset of cardinality κ. In particular, a linear order has cofinality ω if the linear order has a cofinal increasing sequence a0 , a1 , . . . Give a sentence of D which is true in a linear order if and only if the order is either ill-founded or else well-founded and of cofinality ω.

4.5 Connectedness A graph (G, E) is connected if for every two vertices a = b there is a path a1 = a, a2 , . . . , an = b in the graph so that ai Eai+1 for each i = 1, . . . , n − 1. Otherwise the graph is called disconnected. Thus the graph is disconnected if and only if there is a proper subset A of G such that A is closed under E. A graph is disconnected if and only if it satisfies the following sentence: conn : ∃x4 ∃x5 ∀x0 ∃x1 ∀x2 ∃x3 ( =(x2 , x3 )∧ (x0 = x2 → x1 = x3 )∧ (x0 = x4 → x1 = x4 )∧ (x0 = x5 → ¬(x1 = x4 ))∧ ((x1 = x4 ∧ x2 E x0 ) → x3 = x4 ) The explanation is as follows (see Fig. 4.1). The mapping x0 → x1 guesses the set A as the pre-image of x4 . The mapping x2 → x3 is a copy of the mapping x0 → x1 , as witnessed by (x0 = x2 → x1 = x3 ); x4 itself manifests non-emptiness of the guessed set A as witnessed by (x0 = x4 → x1 = x4 ); x5 manifests non-emptiness of the complement of the A as witnessed by (x0 = x5 → ¬(x1 = x4 )). The clause ((x1 = x4 ∧ x2 E x0 ) → x3 E x4 ) guarantees the guessed set is closed under E.

4.6 Natural numbers

57

f x4 x5

Fig. 4.1. Disconnected graph.

Exercise 4.15 Give a sentence of D which is true in a graph if and only if the graph has an infinite clique. (A clique is a subset in which there is an edge between any two distinct vertices.) Exercise 4.16 A cycle of a graph is a finite connected subgraph in which every vertex has degree 2. Give a sentence of D which is true in a finite graph if and only if the graph has a cycle. Exercise 4.17 A path in a graph G is a finite sequence v0 , . . . , vn such that for every i = 0, . . . , n − 1 there is an edge from vi to vi+1 . A Hamiltonian path is a path which passes through every vertex exactly once. Give a sentence of D which is true in a finite graph if and only if the graph has a Hamiltonian path. Exercise 4.18 A graph is 3-colorable if it can be divided into three disjoint parts so that all edges are between elements from different parts. Give a sentence of D which is true in a graph if and only if the graph is 3-colorable. Exercise 4.19 An equivalence relation is an L-structure M for L = {∼} such that ∼M is symmetric (x ∼ y ⇒ y ∼ x), transitive (x ∼ y ∼ z ⇒ x ∼ z), and reflexive (x ∼ x). Give a sentence of D which is true in an equivalence relation if and only if the equivalence relation has infinitely many equivalence classes.

4.6 Natural numbers Let P − be the following first order sentence: ∀x0 (x0 + 0 = 0 + x0 = x0 )∧ ∀x0 ∀x1 (x0 + (x1 + 1) = (x0 + x1 ) + 1)∧ ∀x0 (x0 · 0 = 0 · x0 = 0)∧

Examples

58

∀x0 ∀x1 (x0 · (x1 + 1) = (x0 · x1 ) + x0 )∧ ∀x0 ∀x1 (x0 < x1 ↔ ∃x2 (x0 + (x2 + 1) = x1 )) ∀x0 (x0 > 0 → ∃x1 (x1 + 1 = x0 ))∧ 0 < 1 ∧ ∀x0 (0 < x0 → (1 < x0 ∨ 1 = x0 )) and let N be the following sentence of D, reminiscent of ∞ : ¬P − ∨ ∃x5 ∃x4 ∀x0 ∃x1 ∀x2 ∃x3 (=(x2 , x3 ) ∧ x4 < x5 ∧ ((x0 = x2 ∧ x0 < x5 ) ↔ (x1 = x3 ∧ x1 < x4 ))). The sentence N is of course true in models that do not satisfy the axiom P − . However, in models of N where P − does hold, something interesting happens: the initial segments determined by x4 and x5 are mapped onto each other by the bijection x0 → x1 . Thus such models cannot be isomorphic to (N, +, ·, 0, 1, <), in which all initial segments are finite and of different finite cardinality. Lemma 4.1 If φ is a sentence of dependence logic in the vocabulary of arithmetic, then the following are equivalent: (i) φ is true in (N, +, ·, 0, 1, <); (ii) N ∨ φ is valid in D. Proof Suppose first |= N ∨ φ. Since (N, +, ·, 0, 1, <) |= N , we have necessarily (N, +, ·, 0, 1, <) |= φ. Conversely, suppose (N, +, ·, 0, 1, <) |= φ and let M be arbitrary. If M |= N , then trivially |= N ∨ φ. Suppose then M |= N . Necessarily M |= P − . If M  (N, +, ·, 0, 1, <), we get M |= N , contrary to our assumption. Thus M ∼ = (N, +, ·, 0, 1, <), whence M |= φ.  If Lemma 4.1 is combined with Tarski’s Undefinability of Truth (see Theorem 6.20), we obtain, using φ to denote the G¨odel number of φ according to some obvious G¨odel numbering of sentences of D, the following. Corollary 4.2 The set {φ : φ is valid in D} is non-arithmetical. In particular, there cannot be any effective axiomatization of dependence logic, for then {φ : φ is valid in D} would be recursively enumerable and therefore arithmetical. We return to this important issue in Chapter 7.

4.7 Real numbers

59

4.7 Real numbers Let R F be the first order axiomatization of ordered fields: ∀x0 (x0 + 0 = x0 )∧ ∀x0 ∀x1 (x0 + x1 = x1 + x0 )∧ ∀x0 ∀x1 ∀x2 ((x0 + x1 ) + x2 = x0 + (x1 + x2 ))∧ ∀x0 ∃x1 (x0 + x1 = 0)∧ ∀x0 (x0 · 1 = x0 )∧ ∀x0 ∀x1 (x0 · x1 = x0 · x1 )∧ ∀x0 ∀x1 ∀x2 ((x0 · x1 ) · x2 = x0 · (x1 · x2 ))∧ ∀x0 (x0 = 0 ∨ ∃x1 (x0 · x1 = 1))∧ ∀x0 ∀x1 (x0 · (x1 + x2 ) = x0 · x1 + x0 · x2 )∧ ∀x0 ∀x1 (x0 < x1 → x0 + x2 < x1 + x2 )∧ ∀x0 ∀x1 ((x0 > 0 ∧ x1 > 0) → x0 · x1 > 0). The ordered field (R, +, ·, 0, 1, <) of real numbers is the unique ordered field in which the order is a complete order. The proof of this can be found in standard textbooks on real analysis. Accordingly, let R be the sentence ¬R F ∨ cmpl of D. Exactly as in Lemma 4.1, we have the following. Lemma 4.3 If φ is a sentence of D in the vocabulary of ordered fields, then the following are equivalent: (i) φ is true in (R, +, ·, 0, 1, <); (ii) R ∨ φ is valid in D. This is not as noteworthy as in the case of natural numbers, as the truth of a first order sentence in the ordered field of reals is actually effectively decidable. This is a consequence of the fact, due to Tarski, that this structure admits elimination of quantifiers (see, e.g., ref. [27]). What is noteworthy is that we can add integers to the structure (R, +, ·, 0, 1, <), obtaining the structure (R, +, ·, 0, 1, <, N) with a unary predicate N for the set of natural numbers, making the first order theory of the structure undecidable, and still get a reduction as in Lemma 4.3. To this end, let  N be N 0 ∧ ∀x0 (N x0 → N x0 + 1) ∧ ∀x0 (N x0 → (0 = x0 ∨ 0 < x0 )) ∧ ∀x0 ∀x1 ((N x0 ∧ N x1 ∧ x0 < x1 ) → (x0 + 1 = x1 ∨ x0 + 1 < x1 )).

Examples

60

Let R,N be the sentence ¬R F ∨ cmpl ∨ ¬N of D. Then any structure that is not a model of R,N is isomorphic to (R, +, ·, 0, 1, <, N). Thus we obtain the following lemma easily. Lemma 4.4 If φ is a sentence of D in the vocabulary of ordered fields supplemented by the unary predicate N , then the following are equivalent: (i) φ is true in (R, +, ·, 0, 1, <, N); (ii) R,N ∨ φ is valid in D.

4.8 Set theory The vocabulary of set theory consists of just one binary predicate symbol E. As a precursor to real set theory, let us consider the following simpler situation. We have, in addition to E, two unary predicates R and S. Let θ be the conjunction of the first order sentence: ∀x0 ∀x1 (x0 E x1 → (Rx0 ∧ Sx1 ))∧ ∀x0 (Sx0 → ¬Rx0 ), and the axiom of extensionality: ∀x0 ∀x1 (∀x2 (x2 E x0 ↔ x2 E x1 ) → x0 = x1 ). Canonical examples of models of θ are models of the form (M, ∈, X, P(X )). Indeed, M |= θ if and only if M ∼ = N for some N such that E N = {(a, b) ∈ N 2 : a ∈ R N , b ∈ S N , a ∈ b} and S N ⊆ P(R N ). Let ext : ¬θ ∨ ∃x6 ∀x0 ∃x1 ∀x2 ∃x3 ∀x4 ∃x5 (=(x2 , x3 ) ∧ =(x4 , x5 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x6 → Rx0 ) ∧ ((Sx4 ∧ x0 = x5 ) → (x5 E x4  x1 = x6 ))). The sentence ext is true in a structure M if and only if M ∼ = N for some N such that E N = {(a, b) ∈ N 2 : a ∈ R N , b ∈ S N , a ∈ b} and S N = P(R N ). The explanation is as follows (see Fig. 4.2). The mapping x0 → x1 guesses a set X as the pre-image of x6 . The mapping x2 → x3 is a copy of the mapping x0 → x1 , as witnessed by (x0 = x2 → x1 = x3 ). The clause (x1 = x6 → Rx0 ) makes sure X is a subset of R. The clause ((Sx4 ∧ x0 = x5 ) → (x5 E x4  x1 = x6 )) guarantees the guessed set is not in the set S.

4.8 Set theory

61

S

E

R Fig. 4.2. An element coding a subset.

Lemma 4.5 If φ is a sentence of D in the vocabulary {E, R, S}, then the following are equivalent: (i) φ is true in every model of the form (M, ∈, X, P(X )); (ii) ext ∨ φ is valid in D. The cumulative hierarchy of sets is defined as follows: V0 = ∅, Vα+1 = P(Vα ),  Vν = β<α Vβ for limit ν.

(4.1)

Thus, V1 is the powerset of ∅, i.e. {∅}; V2 is the powerset of {∅}, i.e. {∅, {∅}}; etc. The sets Vn , n ∈ N, are all finite, Vω is countable, and for α > ω the set Vα is uncountable. For more on the cumulative hierarchy, see Section 7.2. Let Z FC ∗ be a large but finite part of the Zermelo–Fraenkel axioms for set theory (see, e.g., ref. [26] for the axioms). It follows from the axioms that every set is in some Vα . Models of the form (Vα , ∈), where α is a limit ordinal, are canonical examples of models of Z FC ∗ . Let set : ¬Z FC ∗ ∨ ∃x6 ∃x7 ∀x0 ∃x1 ∀x2 ∃x3 ∀x4 ∃x5 (=(x2 , x3 ) ∧ ∧ =(x4 , x5 ) ∧ (x0 = x2 → x1 = x3 ) ∧ (x1 = x6 → x0 E x7 ) ∧ (x0 = x5 → (x5 E x4  x1 = x6 ))). Lemma 4.6 If φ is a sentence of D in the vocabulary {E}, then the following are equivalent:

Examples

62

(i) (Vα , ∈) |= φ for all models (Vα , ∈) of Z FC ∗ ; (ii) set ∨ φ is valid in D. In consequence, there are 1 , 2 , 3 and 4 in dependence logic such that (i) (ii) (iii) (iv)

the Continuum Hypothesis holds if and only if 1 is valid in D; the Continuum Hypothesis fails if and only if 2 is valid in D; there are no inaccessible cardinals if and only if 3 is valid in D; there are no measurable cardinals if and only if 4 is valid in D.

These examples show that to decide whether a sentence of D is valid or not is tremendously difficult. One may have to search through the whole set theoretic universe. This is in sharp contrast to first order logic, where to decide whether a sentence is valid or not it suffices to search through finite proofs, i.e. essentially just through natural numbers. By means of the sentence set it is easy to show that for any first order structure M definable in the set theoretical structure (Vω·3 , ∈), which includes virtually all commonly used mathematical structures, and any first order φ, there is a sentence M,φ such that the following are equivalent: (i) φ is true in M; (ii) M,φ is valid in D. Moreover, M,φ can be found effectively on the basis of φ and the defining formula of M. Exercise 4.20 Prove Lemma 4.5. Exercise 4.21 Prove Lemma 4.6. Exercise 4.22 Give a sentence φ of D such that φ has models of all infinite cardinalities, and for all κ ≥ ω, φ has a unique model (up to isomorphism) of cardinality κ if and only if κ is a strong limit cardinal (i.e. λ < κ implies 2λ < κ).

5 Game theoretic semantics

We begin with a review of the well known game theoretic semantics of first order logic (see, e.g., ref. [17]). This is the topic of Section 5.1. There are two ways of extending the first order game to dependence logic. The first, presented in Section 5.2, corresponds to the transition in semantics from assignments to teams. The second game theoretic semantics for dependence logic is closer to the original semantics of independence friendly logic presented in refs. [16] and [19]. In the second game theoretic formulation, the dependence relation =(x0 , . . . , xn ) does not come up as an atomic formula but as the possibility to incorporate imperfect information into the game. A player who aims at securing =(x0 , . . . , xn ) when the game ends has to be able to choose a value for xn only on the basis of what the values of x0 , . . . , xn−1 are. In this sense the player’s information set is restricted to x0 , . . . , xn−1 when he or she chooses xn .

5.1 Semantic game of first order logic The game theoretic semantics of first order logic has a long history. The basic idea is that if a sentence is true, its truth, asserted by us, can be defended against a doubter. A doubter can question the truth of a conjunction φ ∧ ψ by doubting the truth of, say, ψ. He can doubt the truth of a disjunction φ ∨ ψ by asking which of φ and ψ is the one that is true. He can doubt the truth of a negation ¬φ by claiming that φ is true instead of ¬φ. At this point we become the doubter and start questioning why φ is true. This interaction can be formulated in terms of a simple game between two players, I and II. We call them opponents of each other. The opponent of player α is denoted by α ∗ . In the literature these are sometimes called Abelard and Eloise. We go along to the extent that we refer to player I as “he” and to player II as “she.” The players observe a formula φ and an assignment s in the context of a given model M. At the beginning 63

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of the game, player II claims that assignment s satisfies φ in M, and player I doubts this. During the game their roles may change, as we just saw in the case of negation. To keep track of who is claiming what, we use the notation (φ, s, α) for a position in the game. Here α is either I or II. The idea is that α indicates which player is claiming that s satisfies φ in M. Definition 5.1 The semantic game H (φ) of first order logic in a model M is the following game. There are two players, I and II. A position of the game is a triple (ψ, s, α), where ψ is a subformula of φ, s is an assignment, the domain of which contains the free variables of ψ, and α ∈ {I, II}. At the beginning of the game, the position is (φ, ∅, II). The rules of the game are as follows. (i) The position is (t1 = t2 , s, α): if t1M s = t2M s, then player α wins and otherwise the opponent wins. (ii) The position is (Rt1 . . . tn , s, α): if s satisfies Rt1 . . . tn in M, then player α wins, otherwise the opponent wins. (iii) The position is (¬ψ, s, α): the game switches to the position (ψ, s, α ∗ ), where α ∗ is the opponent of α. (iv) The position is (ψ ∨ θ, s, α): the next position is (ψ, s, α) or (θ, s, α), and α decides which. (v) The position is (∃xn ψ, s, α): player α chooses a ∈ M and the next position is (ψ, s(a/xn ), α). The above game is a zero-sum game, i.e. one player’s loss is the other player’s victory. It is also a game of perfect information in the sense that the strategies of both players are allowed to depend on the whole sequence of previous positions. By the Gale–Stewart Theorem [10] all finite zero-sum games of perfect information are determined.1 All the possible positions of this game form, in a canonical way, a tree, which we call the game tree. The game tree for H (φ) starts from the position (φ, ∅, II). Depending on φ, it continues in different ways, as displayed in Figs. 5.1, 5.2, and 5.3. Any (maximal) branch of this tree represents a possible play of the game. We call proper initial segments of plays partial plays. Example 5.2 Consider the game H (∀x0 ∃x1 (x0 E x1 )) in the graph of Fig. 5.4. The tree first splits according to the first move of player I. Since player I can pick any of the four elements, the tree splits into four subtrees. Then

1

The idea of the proof is the following. Suppose I does not have a winning strategy. Player II plays so that always after she has moved player I still does not have a winning strategy. If at some point she has only moves after which player I has a winning strategy, player I had a winning strategy already after the previous move of II; a contradiction.

5.1 Semantic game of first order logic

65

(¬φ, s, α) ↓ (φ, s, α ∗ ) Fig. 5.1. The game tree at negation node.

(φ ∨ ψ, s, α)

(φ, s, α) (ψ, s, α) Fig. 5.2. The game tree at disjunction node.

. . . (φ, s(a/xn ), α)

(∃xn φ, s, α) ↓  (φ, s(a /xn ), α)

(φ, s(a  /xn ), α) . . .

Fig. 5.3. The game tree at quantifier node.

3 p   BB   B    B 0 p  B JJ  B J  B J  B Jp Bp 1 2 Fig. 5.4. A graph for Example 5.2.

player II picks any of the four elements and the tree splits again into four subtrees. Altogether we end up with 16 branches (see Fig. 5.5). The figure shows who wins each play. One can see that whatever player I chooses, player II has a move that guarantees her victory. Inspection of the game tree is vital for success in a game. It is clear that in order to be able to declare victory a player has to have a clear picture in his or her mind what to play in each position. The following concept of strategy is the heart of game theory. It is a mathematically exact concept which tries to capture the idea of a player knowing what to play in each position. Definition 5.3 A strategy of player α in H (φ) is any sequence τ of functions τi defined on the set of all partial plays ( p0 , . . . , pi−1 ) satisfying the following.

Game theoretic semantics

66

2

1

0

3

0

1

2

3

0

1

2

3

0

1

2

3

0

1

2

3

I

II

I

II

II

I

II

II

I

II

I

II

II

II

II

I

Fig. 5.5. A game tree for Example 5.2.

r If p = (φ ∨ ψ, s, α), then τ tells player α which formula to pick, i.e. i−1 τi ( p0 , . . . , pi−1 ) ∈ {0, 1}. If the strategy gives value 0, player α picks the left-hand2 formula φ, and otherwise the right-hand formula ψ. r If p = (∃x φ, s, α), then τ tells player α which element a ∈ M to pick, i.e. i−1 n τi ( p0 , . . . , pi−1 ) ∈ M. We say that player α has used strategy τ in a play of the game H (φ) if in each relevant case player α has used τ to make his or her choice. More exactly, player α has used τ in a play p0 , . . . , pn if the following two conditions hold for all i < n: r if p = (φ ∨ ψ, s, α) and τ ( p , . . . , p ) = 0, then p = (φ, s, α), while i−1 i 0 i−1 i if τi ( p0 , . . . , pi−1 ) = 1, then pi = (ψ, s, α); r if p = (∃x φ, s, α) and τ ( p , . . . , p ) = a, then i−1 m i 0 i−1 pi = (φ, s(a/xm ), α). A strategy τ of player α in the game H (φ) is a winning strategy if player α wins every play in which he or she has used τ . Note that the property of a strategy τ being a winning strategy is defined without any reference to the actual playing of the game. This is not an oversight but an essential feature of the mathematical theory of games. We have reduced 2

One may ask why the values of the strategy are numbers 0 and 1 rather than the formulas themselves. The reason is that the formulas may be one and the same. It is a delicate point whether it then makes any difference which formula is picked. For first order logic there is no difference, but there are extensions of first order logic where a difference emerges.

5.1 Semantic game of first order logic

67

the intuitive act of players choosing their moves to combinatorial properties of some functions. One has to be rather careful in such a reduction. It is possible that the act of playing and handling formulas may use some property of formulas that is intuitively evident but not coded by the mathematical model. One such potential property is the place of a subformula in a formula. We will return to this point in Example 5.9 and the following discussion. Theorem 5.4 Suppose φ is a sentence of first order logic. Then M |=∅ φ in first order logic if and only if player II has a winning strategy in the semantic game H (φ). Proof Suppose M |=∅ φ in first order logic. Consider the following strategy of player II. She maintains the following condition. () If the position is (φ, s, II), then M |=s φ. If the position is (φ, s, I), then M |=s ¬φ. It is completely routine to check that II can actually follow this strategy and win. Note that in the beginning M |=∅ φ, so () holds. For the other direction, suppose player II has a winning strategy τ in the semantic game starting from (φ, ∅, II). It is again completely routine to show the following () If a position (ψ, s, α) is reached in the game, player II using τ , then M |=s ψ if α = II and M |=s ¬ψ if α = I. Since the initial position (φ, ∅, II) is reached at the beginning of the game, we obtain from () the desired conclusion M |=∅ φ.  Exercise 5.1 Consider the game H (∃x0 ∀x1 (x0 = x1 ∨ x0 E x1 )) in the following graph: r r r 1 2 3 Draw the game tree and use it to describe the winning strategy of the player who has it. Exercise 5.2 Consider the game H (∀x0 ∃x1 (x1 E x0 ∧ x0 = x0 )) in the graph of Fig. 5.4. Draw the game tree and use it to describe the winning strategy of the player who has it. Exercise 5.3 Draw the game tree for H (φ), when φ is given by (i) ¬∃x0 P x0 ∧ ¬∃x0 Rx0 ; (ii) ∃x0 (P x0 ∧ Rx0 ); (iii) ∀x0 ∃x1 Rx0 x1 .

Game theoretic semantics

68

Exercise 5.4 Sketch the game tree for H (φ), when φ is given by (i) ∀x0 (P x0 ∨ Rx0 ); (ii) ∀x0 ∃x1 (P x0 ∧ ∃x2 Rx2 x1 ). Exercise 5.5 Sketch the game tree for H (φ), when φ is given by ∃x0 ¬P x0 → ∀x0 (P x0 ∨ Rx0 ). Exercise 5.6 Let L consist of two unary predicates P and R. Let M be an L-structure such that M = {0, 1, 2, 3}, P M = {0, 1, 2}, and R M = {1, 2, 3}. Who has a winning strategy in H (φ) if φ is given by the following: (i) ∃x0 (P x0 ∧ Rx0 ); (ii) ∀x0 ∃x1 ¬(x0 = x1 )? Describe the winning strategy. Exercise 5.7 Let L consist of two unary predicates P and R. Let M be an L-structure such that M = {0, 1, 2, 3}, P M = {0, 1, 2}, and R M = {1, 2, 3}. Who has a winning strategy in H (φ) if φ is given by the following: ∀x0 (P x0 → ∃x1 (¬(x0 = x1 ) ∧ P x0 ∧ Rx1 ))? Describe the winning strategy. Exercise 5.8 Let L consist of two unary predicates P and R. Let M be an L-structure such that M = {0, 1, 2, 3}, P M = {0, 1, 2}, and R M = {1, 2, 3}. Who has a winning strategy in H (φ) if φ is given by the following: ∃x0 (P x0 ∧ ∀x1 ((x0 = x1 ) ∨ P x0 ∨ Rx1 ))? Describe the winning strategy. Exercise 5.9 Suppose M is the binary structure ({0, 1, 2}, R), where R is as in Fig. 5.6. Who has a winning strategy in H (φ) if φ is given by the following: ∀x0 ∃x1 (¬Rx0 x1 ∧ ∀x2 ∃x3 (¬Rx2 x3 ))? Describe the winning strategy. Exercise 5.10 Suppose M is the binary structure ({0, 1, 2}, R), where R is as in Fig. 5.6. Who has a winning strategy in H (φ) if φ is given by the following: ∃x0 ∀x1 (Rx0 x1 ∨ ∃x2 ∀x3 (Rx2 x3 ))? Describe the winning strategy.

5.2 Perfect information game for dependence logic

0

0

1

1

2

2

69

Fig. 5.6. Binary structure.

Exercise 5.11 Show that if τ is a strategy of player II in H (φ) and σ is a strategy of player I in H (φ), then there is one and only one play of H (φ) in which player II has used τ and player I has used σ . (We denote this play by [τ, σ ].) Exercise 5.12 Show that a strategy τ of player II in G(φ) is a winning strategy if and only if player II wins the play [τ, σ ] for every strategy σ of player I.

5.2 Perfect information game for dependence logic In this section we define a game of perfect information, introduced in ref. [41]. This game is very close to our definition of the semantics of dependence logic. In this game the moves are triples (φ, X, d), where X is a team, φ is a formula and d ∈ {0, 1}. If φ is a conjunction and d = 0, we may have an ordered pair of teams. This game has less symmetry than H (φ). On the other hand, the game has formulas of dependence logic as arguments, and D does not enjoy the same kind of symmetry as first order logic. In particular, we cannot let negation correspond to exchanging the roles of the players, as in the case of H (φ), and at the same time have sentences which are neither true nor false. The game we are going to define has two players, I and II. A position in the game is a triple p = (φ, X, d), where φ is a formula, X is a team, the free variables of φ are in dom(X ), and d ∈ {0, 1}. Definition 5.5 Let M be a structure. The game G(φ) is defined by the following inductive definition for all sentences φ of dependence logic. The type of move of each player is determined by the position as follows. (M1) The position is (φ, X, 1), and φ = φ(xi1 , . . . , xik ) is of the form t1 = t2 or of the form Rt1 . . . tn . Then the game ends. Player II wins if (∀s ∈ X )(M |= φ(s(xi1 ), . . . , s(xik ))).

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Game theoretic semantics

Otherwise player I wins. (M2) The position is (φ, X, 0), and φ = φ(xi1 , . . . , xik ) is of the form t1 = t2 or of the form Rt1 . . . tn . Then the game ends. Player II wins if (∀s ∈ X )(M |= φ(s(xi1 ), . . . , s(xik ))). Otherwise player I wins. (M3) The position is (=(t1 , . . . , tn ), X, 1). Then the game ends. Player II wins if M |= X =(t1 , . . . , tn ). Otherwise player I wins. (M4) The position is (=(t1 , . . . , tn ), X, 0). Then the game ends. Player II wins if X = ∅. Otherwise player I wins. (M5) The position is (¬φ, X, 1). The game continues from the position (φ, X, 0). (M6) The position is (¬φ, X, 0). The game continues from the position (φ, X, 1). (M7) The position is (φ ∨ ψ, X, 1). Now player II chooses Y and Z such that X = Y ∪ Z , and we move to position (φ ∨ ψ, (Y, Z ), 1). Then player I chooses whether the game continues from position (φ, Y, 1) or (ψ, Z , 1). (M8) The position is (φ ∨ ψ, X, 0). Now player I chooses whether the game continues from position (φ, X, 0) or (ψ, X, 0). (M9) The position is (∃xn φ, X, 1). Now player II chooses F : X → M, and then the game continues from the position (φ, X (F/xn ), 1). (M10) The position is (∃xn φ, X, 0). Now the game continues from the position (φ, X (M/xn ), 0). At the beginning of the game, the position is (φ, {∅}, 1). Note that case (M8) generates two rounds for the game: during the first round, player II makes a choice for Y and Z . We call this round a half-round. During the next round, player I makes a choice between them. Thus, after the position (φ ∨ ψ, X, 1) there is the position (φ ∨ ψ, (Y, Z ), 1), from which the game then proceeds to either (φ, Y, 1) or (ψ, Z , 1). Note also that player II has something to do only in the cases (M7) and (M9). Likewise, player I has something to do only in cases (M7) and (M8). Otherwise the game goes on in a determined way with no interaction from the players. All the possible positions of this game form a tree, just as in the case of the game H (φ). The tree for G(φ) starts from the position (φ, {∅}, 1). Depending on φ it continues in different ways as displayed in Figs. 5.7, 5.8 and 5.9. Note that player II has always a winning strategy in a position of the form (φ, ∅, d). Example 5.6 Suppose M has at least two elements. Player I has a winning strategy in G(∀x0 ∃x1 (=(x1 ) ∧ x0 = x1 )). The game tree is in Fig 5.10. Note

5.2 Perfect information game for dependence logic

(¬φ, X, 1) ↓ (φ, X, 0)

71

(∃xn φ, X, 0) ↓ (φ, X (M/xn ), 1)

(¬φ, X, 0) ↓ (φ, X, 1)

Fig. 5.7. Game trees do not split.

(φ ∨ ψ, X, 0)

(φ, X, 0)

(ψ, X, 0)

Fig. 5.8. The game tree splits into two.

(φ ∨ ψ, X, 1)

↓ 

. . . (φ ∨ ψ, (Y, Z ), 1)



(φ ∨ ψ, (Y , Z ), 1)

(φ ∨ ψ, (Y  , Z  ), 1) . . .

(∃xn φ, X, 1) . . . (φ, X (F/xn ), 1)

↓



(φ, X (F  /xn ), 1) . . .

(φ, X (F /xn ), 1)

Fig. 5.9. The game tree splits into many trees.

that when (M9) is applied, the tree splits into as many branches as there are functions F. At the end of the game, the winner is decided on the basis of (M1)–(M4). We define now what we mean by a strategy in the game G(φ). Definition 5.7 A strategy of player II in G(φ) is any sequence τ of functions τi defined on the set of all partial plays ( p0 , . . . , pi−1 ) satisfying the following: r if p = (φ ∨ ψ, X, 1), then τ tells player II how to cover X with two sets, i−1 one corresponding to φ and the other to ψ, i.e. τi ( p0 , . . . , pi−1 ) = (Y, Z ) such that X = Y ∪ Z ; r if p = (∃x φ, X, 1), then τ tells player II how to supplement X , i.e. i−1 n τi ( p0 , . . . , pi−1 ) is a function F : X → M. We say that player II has used strategy τ in a play of the game G(φ) if in the cases (M7) and (M9) player II has used τ to make the choice. More exactly, player II has used τ in a play p0 , . . . , pn if the following two conditions hold for all i < n:

Game theoretic semantics

72

(¬∃x0 ¬∃x1 (=(x1 ) ∧ x0 = x1 ), X, 1) ↓ (M5) (∃x0 ¬∃x1 (=(x1 ) ∧ x0 = x1 ), X, 0) ↓ (M10) (¬∃x1 (=(x1 ) ∧ x0 = x1 ), X (M/x0 ), 0) ↓ (M5) (∃x1 (=(x1 ) ∧ x0 = x1 ), X (M/x0 ), 1) ···

↓ (M9)

((=(x1 ) ∧ x0 = x1 ), X (M/x0 )(F/x1 ), 1)

···

↓ (M5) ···

((¬ =(x1 ) ∨ ¬x0 = x1 ), X (M/x0 )(F/x1 ), 0)

(¬ =(x1 ), X (M/x0 )(F/x1 ), 0) ↓ (=(x1 ), X (M/x0 )(F/x1 ), 1)

···

(M8) (¬x0 = x1 , X (M/x0 )(F/x1 ), 0) ↓ (M6) (x0 = x1 , X (M/x0 )(F/x1 ), 1)

Fig. 5.10. The game tree in Example 5.6.

r if pi−1 = (φ ∨ ψ, X, 1) and τi ( p0 , . . . , pi−1 ) = (Y, Z ), then pi = (φ ∨ ψ, (Y, Z ), 1); r if p = (∃x φ, X, 1) and τ ( p , . . . , p ) = F, then p = (φ, X (F/x ), i−1 n i 0 i−1 i n 1). A strategy of player I in G(φ) is any sequence σ of functions σi defined on the set of all partial plays p0 , . . . , pi−1 satisfying the following. r If p = (φ ∨ ψ, X, 0), then σ tells player I which formula to pick, i.e. i−1 σi ( p0 , . . . , pi−1 ) ∈ {0, 1}. If the strategy gives value 0, player I picks the left-hand formula φ, and otherwise he chooses the right-hand formula ψ. r If p = (φ ∨ ψ, (Y, Z ), 1), then σ ( p , . . . , p ) ∈ {0, 1}. i−1 i 0 i−1 We say that player I has used strategy σ in a play of the game G(φ) if in the cases (M7) and (M8) player I has used σ to make the choice. More exactly, I has used σ in a play p0 , . . . , pn if the following two conditions hold: r if p = (φ ∨ ψ, X, 0) and σ ( p , . . . , p ) = 0, then p = (φ, X, 0), and i−1 i 0 i−1 i if σi ( p0 , . . . , pi−1 ) = 1, then pi = (ψ, X, 0).

5.2 Perfect information game for dependence logic

73

r if p = (φ ∨ ψ, (Y, Z ), 1), then p = (φ, Y, 0) if σ ( p , . . . , p ) = 0 and i−1 i i 0 i−1 pi = (ψ, Z , 0) if σi ( p0 , . . . , pi−1 ) = 1. A strategy of player α in the game G(φ) is a winning strategy if player α wins every play in which he or she has used the strategy. Theorem 5.8 M |= φ if and only if player II has a winning strategy in G(φ). Proof Assume first M |= φ. We describe a winning strategy of player II in G(φ). Player II maintains in G(φ) the condition that if the position (omitting half-rounds) is (ψ, X, d), then (ψ, X, d) ∈ T . We prove this by induction on φ. (S1) Position (φ, X, 1), where φ is t = t  or Rt1 . . . tn . Since (φ, X, 1) ∈ T , player II wins, by Definition 5.5 (M1). (S2) Position (φ, X, 0), where φ is t = t  or Rt1 . . . tn . Since (φ, X, 0) ∈ T , player II wins, by Definition 5.5 (M2). (S3) Position (=(t1 , . . . , tn ), X, 1). Since (=(t1 , . . . , tn ), X, 1) ∈ T , player II wins, by definition. (S4) Position (=(t1 , . . . , tn ), X, 0). Since (=(t1 , . . . , tn ), X, 0) ∈ T , we have X = ∅, and therefore II wins by definition. (S5) Position (¬φ, X, 1). Since (¬φ, X, 1) ∈ T , we have (φ, X, 0) ∈ T . Thus II can play this move according to her plan. (S6) Position (¬φ, X, 0). Since (¬φ, X, 0) ∈ T , we have (φ, X, 1) ∈ T . Thus II can play this move according to her plan. (S7) Position (φ ∨ ψ, X, 0). We know (φ ∨ ψ, X, 0) ∈ T and therefore both (φ, X, 0) ∈ T and (ψ, X, 0) ∈ T . Thus, whether the game proceeds to (φ, X, 0) or (ψ, X, 0), player II maintains her plan. (S8) Position (φ ∨ ψ, X, 1). We know (φ ∨ ψ, X, 1) ∈ T and hence (φ, Y, 1) ∈ T and (ψ, Z , 1) ∈ T for some Y and Z with X = Y ∪ Z . Thus we can let player II play the ordered pair (Y, Z ). After this halfround player I wants the game to proceed either to (φ, Y, 1) or to (ψ, Z , 1). In either case player II can fulfil her plan, as both (φ, Y, 1) ∈ T and (ψ, Z , 1) ∈ T . (S9) Position (∃xn φ, X, 1). Thus there is F : X → M such that the triple (φ, X (F/xn ), 1) is in T . Player II can now play the function F, for in the resulting position (φ, X (F/xn ), 1) she can maintain the condition (φ, X (F/xn ), 1) ∈ T . (S10) Position (∃xn φ, X, 0). We know (φ, X (M/xn ), 0) ∈ T . But the triple (φ, X (M/xn ), 0) is the next position, so II can maintain her plan.

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Game theoretic semantics

For the other direction, we assume player II has a winning strategy τ in G(φ) and use this to show M |= φ. We prove by induction on φ that if II is using τ and a position (ψ, X, d) is reached, then (ψ, X, d) ∈ T . This gives the desired conclusion as the initial position (φ, {∅}, 1) is trivially reached. (S1 ) Position (φ, X, 1), where φ is t = t  or Rt1 . . . tn . Since II has been playing her winning strategy, (φ, X, 1) ∈ T by Definition 5.5 (M1).  (S2 ) Position (φ, X, 0), where φ is t = t  or Rt1 . . . tn . Since II has been playing her winning strategy, (φ, X, 0) ∈ T by Definition 5.5 (M2). (S3 ) Position (=(t1 , . . . , tn ), X, 1). Since II has been playing her winning strategy, (=(t1 , . . . , tn ), X, 1) ∈ T , by definition.  (S4 ) Position (=(t1 , . . . , tn ), X, 0). Since II has been playing her winning strategy, X = ∅, and therefore (=(t1 , . . . , tn ), X, 0) ∈ T by definition. (S5 ) Position (¬φ, X, 1). The game continues, still following τ , to the position (φ, X, 0). By the induction hypothesis, (φ, X, 0) ∈ T , and therefore (¬φ, X, 1) ∈ T . (S6 ) Position (¬φ, X, 0). The game continues, still following τ , to the position (φ, X, 1). By the induction hypothesis, (φ, X, 1) ∈ T , and therefore (¬φ, X, 0) ∈ T .  (S7 ) Position (φ ∨ ψ, X, 0). To prove (φ ∨ ψ, X, 0) ∈ T , we need both (φ, X, 0) ∈ T and (ψ, X, 0) ∈ T . Let us try to prove (φ, X, 0) ∈ T . Since II is following a winning strategy, we can let the game proceed to position (φ, X, 0). By the induction hypothesis, (φ, X, 0) ∈ T . The same argument gives (ψ, X, 0) ∈ T . Thus we have proved (φ ∨ ψ, X, 0) ∈ T . (S8 ) Position (φ ∨ ψ, X, 1). To get (φ ∨ ψ, X, 1) ∈ T we need (φ, Y, 1) ∈ T and (ψ, Z , 1) ∈ T for some Y and Z with X = Y ∪ Z . Indeed, the winning strategy τ gives an ordered pair (Y, Z ) with X = Y ∪ Z . Let us try to prove (φ, Y, 1) ∈ T . Since II is following a winning strategy, we can let the game proceed to position (φ, Y, 1). By the induction hypothesis, (φ, Y, 1) ∈ T . The same argument gives (ψ, Z , 1) ∈ T . Thus we have proved (φ ∨ ψ, X, 1) ∈ T .  (S9 ) Position (∃xn φ, X, 1). The strategy τ gives F : X → M such that II has a winning strategy in position (φ, X (F/xn ), 1). By the induction hypothesis, (φ, X (F/xn ), 1) ∈ T . Hence (∃xn φ, X, 1) ∈ T . (S10 ) Position (∃xn φ, X, 0). The game continues, still following τ , to the position (φ, X (M/xn ), 0). By the induction hypothesis the triple (φ, X (M/xn ), 0) is in T , and therefore (∃xn φ, X, 0) ∈ T .  Theo Janssen [24] has pointed out the following example.

5.2 Perfect information game for dependence logic

75

Table 5.1.

(

=

(

(= (x0 ) ∨ ¬x0 = x1 ) x0 ) ∨ ¬ x0 =

x1

)

1

2

3

4

10

11

5

6

7

8

9

Example 5.9 The sentence ∀x0 ∃x1 ((=(x1 ) ∧ ¬x0 = x1 ) ∨ (=(x1 ) ∧ ¬x0 = x1 ))

(5.1)

is true in the natural numbers. The trick is the following: for s ∈ {∅}(N/x0 ) let F(s) ∈ {0, 1} be such that F(s) = s(x0 ). The team {∅}(N/x0 )(F/x1 ) is a subset of the union of {∅}(N/x0 )(F0 /x1 ) and {∅}(N/x0 )(F1 /x1 ), where F0 is the constant function 0 and F1 is the constant function 1. Both parts satisfy (=(x1 ) ∧ ¬x0 = x1 ). The above example shows that when we play a game that follows the structure of a formula, we may have to take the formula structure into the game. To accomplish this, and in order to be sufficiently precise, we identify formulas with finite strings of symbols. Variables xn are treated as separate symbols. The other symbols are the symbols of the vocabulary, =, ), (, ¬, ∧, ∃, and the comma. Each string S has a length, which we denote by len(S). We number the symbols in a formula with positive integers starting from the left, as in Table 5.1 If the nth symbol of φ starts a string which is a subformula of φ, we denote the subformula by (φ, n). Thus, every subformula of φ is of the form (φ, n) for some n and some may occur with several n. In the case that φ is the formula in Eq. (5.1), the subformula (=(x1 ) ∧ ¬x0 = x1 ) occurs as both (φ, 6) and (φ, 18). Note that (i) if (φ, n) = ¬ψ, then (φ, n + 1) = ψ; (ii) if (φ, n) = (ψ ∨ θ ), then (φ, n + 1) = ψ and (φ, n + 2 + len(ψ)) = θ; (iii) if (φ, n) = ∃xm ψ, then (φ, n + 2) = ψ. We let G place (φ) be the elaboration of the game G(φ) in which the rules are the same as in G(φ) but the positions are of the form (ψ, n, X, d), and it is assumed all the time that (φ, n) = ψ. Thus the formula ψ could be computed from the number n, and it is mentioned in the position only for the sake of clarity.

Game theoretic semantics

76

Definition 5.10 Let M be a structure. The game G place (φ) is defined by the following inductive definition for all sentences φ of dependence logic. The type of the move of each player is determined by the position as follows. (M1 ) The position is (φ, n, X, 1) and φ = φ(xi1 , . . . , xik ) is of the form t1 = t2 or of the form Rt1 . . . tn . Then the game ends. Player II wins if (∀s ∈ X )(M |= φ(s(xi1 ), . . . , s(xik ))). Otherwise player I wins. (M2 ) The position is (φ, n, X, 0) and φ = φ(xi1 , . . . , xik ) is of the form t1 = t2 or of the form Rt1 . . . tn . Then the game ends. Player II wins if 

(∀s ∈ X )(M |= φ(s(xi1 ), . . . , s(xik ))). Otherwise player I wins. (M3 ) The position is (=(t1 , . . . , tn ), m, X, 1). Then the game ends. Player II wins if M |= X =(t1 , . . . , tn ). Otherwise player I wins.  (M4 ) The position is (=(t1 , . . . , tn ), m, X, 0). Then the game ends. Player II wins if X = ∅. Otherwise player I wins. (M5 ) The position is (¬φ, n, X, 1). The game continues from the position (φ, n + 1, X, 0).  (M6 ) The position is (¬φ, n, X, 0). The game continues from the position (φ, n + 1, X, 1). (M7 ) The position is (φ ∨ ψ, n, X, 0). Now player I chooses whether the game continues from position (φ, n + 1, X, 0) or (ψ, n + 2 + len(φ), X, 0). (M8 ) The position is (φ ∨ ψ, n, X, 1). Now player II chooses X 0 and X 1 such that X = X 0 ∪ X 1 , and we move to position (φ ∨ ψ, n, (X 0 , X 1 ), 1). Then player I chooses whether the game continues from position (φ, n + 1, X 0 , 1) or (ψ, n + 2 + len(φ), X 1 , 1). (M9 ) The position is (∃xn φ, m, X, 1). Now player II chooses F : X → M and then the game continues from the position (φ, m + 2, X (F/xn ), 1). (M10 ) The position is (∃xn φ, m, X, 0). Now the game continues from the position (φ, m + 2, X (M/xn ), 0). At the beginning of the game, the position is (φ, 1, {∅}, 1).

5.2 Perfect information game for dependence logic

77

The following easy observation shows that coding the location of the subformula into the game makes no difference. However, we shall use G place (φ) in the proof of Theorem 5.16. Proposition 5.11 If player II has a winning strategy in G(φ), she has a winning strategy in G place (φ), and vice versa. Proof Assume II has a winning strategy τ in G(φ). We describe her winning strategy in G place (φ). While playing G place (φ), she also plays G(φ), maintaining the condition that if the position in G place (φ) is (φ, n, X, d), then the position in G(φ) is (φ, X, d) while she uses τ in G(φ). (i) Position is (φ, n, X, d), where φ is t = t  or Rt1 . . . tn . Player II wins since, by assumption, she wins G(φ) in position (φ, X, d). (ii) Position is (=(t1 , . . . , tn ), n, X, d). Player II wins since, by assumption, she wins G(φ) in position (=(t1 , . . . , tn ), X, d). (iii) Position is (¬φ, n, X, d). Player II moves to the position (φ, n + 1, X, 1 − d) in G place (φ) and to the position (φ, X, 1 − d) in G(φ). Her strategy is still valid. (iv) Position is ((φ ∨ ψ), n, X, 0). Player II proceeds, according to the choice of player I, to the position (φ, n + 1, X, 0) or to the position (ψ, n + 2 + len(φ), X, 0) in G place (φ) and, respectively, to the position (φ, X, 0) or to the position (ψ, X, 0) in G(φ). Whichever way the game proceeds, her strategy is still valid. (v) Position is ((φ ∨ ψ), n, X, 1). We know that II is using τ in G(φ), so in the position ((φ ∨ ψ), X, 1) ∈ T she has Y and Z with X = Y ∪ Z , and she can still win with τ from positions (φ, Y, 1) and (ψ, Z , 1). Thus we let player II play the ordered pair (Y, Z ). After this half-round, player I wants the game to proceed either to (φ, n + 1, Y, 1) or to (ψ, n + 2 + len(φ), Z , 1). In either case player II can maintain her plan. (vi) Position is (∃xn φ, m, X, 1). The strategy τ gives player II a function F : X → M and the game G(φ) proceeds to the position (φ, X (F/xn ), 1). We let II play this function F. The game G place (φ) proceeds to the position (φ, m + 2, X (F/xn ), 1) and II maintains her plan. (vii) Position is (∃xn φ, m, X, 0). Player II proceeds to the position (φ, ¥m + 2, X (M/xn ), 0) in G place (φ) and to the respective position (φ, X (M/xn ), d) in G(φ). Player II maintains her plan. For the other direction, assume II has a winning strategy τ in G place (φ). We describe her winning strategy in G(φ). While playing G(φ) she also plays G place (φ), maintaining the condition that if the position in G(φ) is (φ, X, d), then the position in G(φ) is (φ, n, X, d) for some n while she uses τ in G place (φ).

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Game theoretic semantics

(i) Position is (ψ, X, d), where ψ is atomic. Player II wins since, by assumption she wins G place (φ) in position (ψ, n, X, d) for some n. (ii) Position in G place (φ) is (¬φ, X, d) and in G place (φ) it is (¬φ, n, X, d). Player II moves to the position (φ, n + 1, X, 1 − d) in G place (φ) and to the position (φ, X, 1 − d) in G(φ). Her strategy is still valid. (iii) Position is ((φ ∨ ψ), X, 0) and in G place (φ) it is ((φ ∨ ψ), n, X, 0). Player II moves to the position (φ, X, 0) or to the position (ψ, X, 0) in G place (φ) and, respectively, to the position (φ, n + 1, X, 0) or to the position (ψ, n + 2 + len(φ), X, 0) in G(φ). Whichever way the game proceeds, her strategy is still valid. (iv) Position is ((φ ∨ ψ), X, 1) and in G place (φ) it is ((φ ∧ ψ), n, X, 1). We know that II is using τ in G place (φ) so in the position ((φ ∨ ψ), n, X, 1) ∈ T she has Y and Z with X = Y ∪ Z , and she can still win with τ from positions (φ, n + 1, Y, 1) and (ψ, n + 2 + len(φ), Z , 1). Thus we let player II play the ordered pair (Y, Z ). After this half-round, player I wants the game to proceed either to (φ, Y, 1) or to (ψ, Z , 1). Player II moves in G place (φ), respectively, to (φ, n + 1, Y, 1) or to (ψ, n + 2 + len(φ), Z , 1). In either case, player II can maintain her plan. (v) Position is (∃xn φ, X, 1) and in G place (φ) it is (∃xn φ, m, X, 1). The strategy τ gives player II a function F : X → M and the game G place (φ) proceeds to the position (φ, m + 2, X (F/xn ), 1). We let II play this function F. The game G(φ) proceeds to the position (φ, X (F/xn ), 1), and II maintains her plan. (vi) Position is (∃xn φ, X, 0) and in G place (φ) it is (∃xn φ, m, X, 0). Player II proceeds to the position (φ, ¥m + 2, X (M/xn ), 0) in G place (φ) and to the  position (φ, X (M/xn ), d) in G(φ). Player II maintains her plan. Lemma 5.12 Suppose player II uses the strategy τ in G place (φ) and the game reaches a position (ψ, n, X, d). Then X and d are uniquely determined by τ and n. Proof Let k0 , . . . , km be the unique sequence of numbers such that if we denote (φ, ki ) by φi , then φ0 = φ, φi+1 is an immediate subformula of φi , and φm = ψ. This sequence is uniquely determined by the number n. During the game that ended in (ψ, n, X, d), the positions (omitting halfrounds) were (φi , ki , X i , di ), i = 0, . . . , m. We know that φ0 = φ, X 0 = {∅}, and d0 = 1. If φi = ¬φi+1 , then necessarily X i+1 = X i and di+1 = 1 − di . If φi+1 is a conjunct of φi , and di = 1, then X i+1 = X i and di+1 = 1. If φi+1 is a conjunct of φi = ψ ∧ θ , and di = 0, then τ determines, on the basis of (φ j , k j , X j , d j ), j = 0, . . . , i, two sets Y and Z such that X i = Y ∪ Z . Player I chooses whether X i+1 = Y or X i+1 = Z . The result is completely

5.2 Perfect information game for dependence logic

79

determined by whether ki+1 = ki + 1 or ki+1 = ki + 2 + len(ψ). If φi = ∃xn φi+1 and di = 0, then X i+1 = X i (M/xn ) and di+1 = di , both uniquely determined by X i and di . If φi = ∃xn φi+1 and di = 1, then τ determines F : X → M on the basis of (φ j , k j , X j , d j ), j = 0, . . . , i. Then X i+1 = X i (F/xn ) and di = 1, again uniquely determined by τ and X i . 

Exercise 5.13 Draw the game tree for G(φ), when φ is given by (i) ¬∃x0 P x0 ∧ ¬∃x0 Rx0 , (ii) ∃x0 (P x0 ∧ Rx0 ). Exercise 5.14 Draw the game tree for G(φ), when φ is given by (i) ∀x0 (P x0 ∨ Rx0 ), (ii) ∀x0 ∃x1 (P x0 ∧ ∃x2 Rx2 x1 ). Exercise 5.15 Draw the game tree for G(φ), when φ is given by ∃x0 ¬P x0 → ∀x0 (P x0 ∨ Rx0 ). Exercise 5.16 Let L consist of two unary predicates P and R. Let M be an L-structure such that M = {0, 1, 2, 3}, P M = {0, 1, 2}, and R M = {1, 2, 3}. Who has a winning strategy in G(φ) if φ is given by the following: (i) ∃x0 (P x0 ∧ Rx0 ), (ii) ∀x0 ∃x1 ¬(x0 = x1 )? Describe the winning strategy. Exercise 5.17 Let L consist of two unary predicates P and R. Let M be an L-structure such that M = {0, 1, 2, 3}, P M = {0, 1, 2}, and R M = {1, 2, 3}. Who has a winning strategy in G(φ) if φ is given by the following ∀x0 (P x0 → ∃x1 (¬(x0 = x1 ) ∧ P x0 ∧ Rx1 ))? Describe the winning strategy. Exercise 5.18 Let L consist of two unary predicates P and R. Let M be an L-structure such that M = {0, 1, 2, 3}, P M = {0, 1, 2}, and R M = {1, 2, 3}. Who has a winning strategy in G(φ) if φ is given by the following: ∃x0 (P x0 ∧ ∀x1 ((x0 = x1 ) ∨ P x0 ∨ Rx1 ))? Describe the winning strategy. Exercise 5.19 Use the game tree to analyze the formula in Eq. (5.1).

80

Game theoretic semantics

Exercise 5.20 Suppose M is the binary structure ({0, 1, 2}, R), where R is as in Fig. 5.6. Who has a winning strategy in G(φ) if φ is given by the following: ∀x0 ∃x1 (¬Rx0 x1 ∧ ∀x2 ∃x3 (=(x2 , x3 ) ∧ ¬Rx0 x3 ))? Describe the winning strategy. Exercise 5.21 Suppose M is the binary structure ({0, 1, 2}, R), where R is as in Fig. 5.6. Who has a winning strategy in G(φ) if φ is given j by the following: ∃x0 ∀x1 (Rx0 x1 ∨ ∃x2 ∀x3 (=(x2 , x3 ) → Rx2 x3 ))? Describe the winning strategy. Exercise 5.22 Show that if τ is a strategy of player II in G(φ) and σ is a strategy of player I in G(φ), then there is one and only one play of G(φ) in which player II has used τ and player I has used σ . (We denote this play by [τ, σ ].) Exercise 5.23 Show that a strategy τ of player II in G(φ) is a winning strategy if and only if player II wins the play [τ, σ ] for every strategy σ of player I.

5.3 Imperfect information game for dependence logic The semantics of dependence logic can also be defined by means of a simpler game. In this case, however, we have to put a uniformity restriction on strategies in order to get the correct truth definition. The restriction has the effect of making the game a game of partial information. As in Section 5.2, we pay attention to where a subformula occurs in a formula. This is taken care of by the parameter n in Definition 5.13. It should be borne in mind that disjunctions are assumed to have brackets around them, as in (ψ ∨ θ). Then if this formula is (φ, n), we can infer that ψ is (φ, n + 1) and θ is (φ, n + 2 + len(ψ)). Definition 5.13 Let φ be a sentence of dependence logic. The semantic game H (φ) in a model M is the following game: there are two players, I and II. A position of the game is a quadruple (ψ, n, s, α), where ψ is (φ, n), s is an assignment, the domain of which contains the free variables of ψ, and α ∈ {I, II}. At the beginning of the game the position is (φ, 1, ∅, II). The rules of the game are as follows. (i) The position is (t1 = t2 , n, s, α): if t1M s = t2M s, then player α wins and otherwise the opponent wins.

5.3 Imperfect information game for dependence logic

81

(¬φ, n, s, α) ↓ (φ, n + 1, s, α ∗ ) Fig. 5.11. The game tree of H (φ) at negation node.

(φ ∨ ψ, n, s, α)

(φ, n + 1, s, α) (ψ, n + 2 + len(φ), s, α) Fig. 5.12. The game tree of H (φ) at disjunction node.

(ii) The position is (Rt1 . . . tm , n, s, α): if s satisfies Rt1 . . . tm in M, then player α wins, otherwise the opponent wins. (iii) The position is (=(t1 , . . . , tm ), n, s, α): player α wins. (iv) The position is (¬φ, n, s, α): the game switches to the position (φ, n + 1, s, α ∗ ), where α ∗ is the opponent of α. (v) The position is ((ψ ∨ θ ), n, s, α): the next position is (ψ, n + 1, s, α) or (θ, n + 2 + len(ψ), s, α), and α decides which. (vi) The position is (∃xm φ, n, s, α): player α chooses a ∈ M and the next position is (φ, n + 2, s(a/xm ), α). Thus (=(t1 , . . . , tn ), n, s, α) is a safe haven for α. Note that the game is a determined zero-sum game of perfect information. However, we are not really interested in who has a winning strategy in this determined game, but in who has a winning strategy with extra uniformity, as defined below. The uniformity requirement in effect makes the game into a non-determined game of imperfect information. The concepts of a game tree, play, and partial play are defined for this game exactly as for the game G(φ). Definition 5.14 A strategy of player α in H (φ) is any sequence τ of functions τi defined on the set of all partial plays ( p0 , . . . , pi−1 ) satisfying the following. r If p = ((φ ∨ ψ), n, s, α), then τ tells player α which formula to pick, i.e. i−1 τi ( p0 , . . . , pi−1 ) ∈ {n + 1, n + 2 + len(φ)}. If the strategy gives the lower value, player α picks the left-hand formula φ, and otherwise chooses the right-hand formula ψ. r If p = (∃x φ, n, s, α), then τ tells player α which element a ∈ M to pick, i−1 m i.e. τi ( p0 , . . . , pi−1 ) ∈ M.

82

Game theoretic semantics

(∃xm φ, n, s, α) ↓

. . . (φ, n + 2, s(a/xm ), α) (φ, , n + 2, s(a  /xm ), α) (φ, n + 2, s(a  /xm ), α) . . . Fig. 5.13. The game tree of H (φ) at quantifier node.

We say that player α has used strategy τ in a play of the game H (φ) if in each relevant case player α has used τ to make his or her choice. More exactly, player α has used τ in a play p0 , . . . , pn if the following two conditions hold for all i < n: r if p = ((φ ∨ ψ), m, s, α) and τ ( p , . . . , p ) = m + 1, then p = i−1 i 0 i−1 i (φ, m + 1, s, α), while if τi ( p0 , . . . , pi−1 ) = m + 2 + len(φ), then pi = (ψ, m + 2 + len(φ), s, α); r if p = (∃x φ, m, s, α) and τ ( p , . . . , p ) = a, then p = (φ, m + i−1 k i 0 i−1 i 2, s(a/xk ), α). A strategy of player α in the game H (φ) is a winning strategy if player α wins every play in which she has used the strategy. Definition 5.15 We call a strategy τ of player II in the game H (φ) uniform if the following condition holds. Suppose ( (φ, m), m, s, II) and ( (φ, m), m, s  , II) are two positions arising in the game when II has played according to τ . Moreover we assume that (φ, m) is =(t1 , . . . , tn ). Then if s and s  agree about the values of t1 , . . . , tn−1 , they agree about the value of tn . Theorem 5.16 Suppose φ is a sentence of dependence logic. Then M |={∅} φ if and only if player II has a uniform winning strategy in the semantic game H (φ). Proof Suppose M |= X φ. Let τ be a winning strategy of II in G place (φ). Consider the following strategy of player II. She keeps playing G place (φ) as an auxiliary game such that if she is in a position (φ, n, X, d) in G place (φ) and has just moved in the semantic game, the following holds. () Suppose the position is (φ, n, s, α). Then II is in a position (φ, n, X, d), playing τ , in G place (φ) and s ∈ X . If α = II, then d = 1. If α = I, then d = 0. Let us check that II can actually follow this strategy and win. In the beginning M |={∅} φ, so () holds. (i) φ is t1 = t2 or Rt1 . . . tn . If α = II, s satisfies φ in M. So II wins. If α = I, s does not satisfy φ in M, and again II wins.

5.3 Imperfect information game for dependence logic

83

(ii) φ is =(t1 , . . . , tn ). If α = II, then II wins by definition. On the other hand, s ∈ X , so X = ∅, and we must have α = II. (iii) φ is ¬ψ and the position in G place (φ) is (¬ψ, n, X, d). By the rules of G place (φ), the next position is (ψ, n + 1, X, 1 − d). So the game can proceed to position (ψ, n + 1, s, α ∗ ) and II maintains (). (iv) φ is (ψ ∨ θ) and the position in G place (φ) is ((ψ ∨ θ ), n, X, d). Suppose α = I and d = 0. Then both (ψ, n + 1, X, d) and (θ, n + 2 + len(ψ), X, d) are possible positions in G place (φ) while II uses τ . The next position is (ψ, n + 1, s, 0) or (θ, n + 2 + len(ψ0 ), s, 0), and I chooses which. Condition () remains valid, whichever he chooses. Suppose then α = II and d = 1. Strategy τ gives X 0 and X 1 such that X = X 0 ∪ X 1 and II wins with τ both in the position (ψ, n + 1, X 0 , 1) and in (θ, n + 2 + len(ψ0 ), X 1 , 1). Since s ∈ X , we have either s ∈ X 0 or s ∈ X 1 . Let us say s ∈ X 0 . We let I play ψ in G place (φ). The game G place (φ) proceeds to (ψ, n + 1, X 0 , 1). We let II play in H (φ) the sentence ψ. Condition () remains valid. The situation is similar if s ∈ X 1 . (v) φ is ∃xψ. We leave this as an exercise. We claim that the strategy is uniform. Suppose s and s  are assignments arising from the game when II plays the above strategy and the game ends in the same dependence formula =(t1 , . . . , tn ). Let the ending positions be ( (φ, n), s, α) and ( (φ, n), s  , α). Since II wins, α = II. When the game ended, player II had reached the position (=(t1 , . . . , tn ), n, X, 1) on one hand and the position (=(t1 , . . . , tn ), n, X  , 1) on the other hand in G place (φ) playing, τ . By Lemma 5.12, X = X  . Suppose s and s  agree about the values of t1 , . . . , tn−1 . Since II wins in the position (=(t1 , . . . , tn ), n, X, 1) and s, s  ∈ X , it follows that s and s  agree about the value of tn . This strategy gives one direction of the theorem. For the other direction, suppose player II has a uniform winning strategy τ in the semantic game starting from (φ, 1, ∅, II). Let X n be the set of s such that ( (φ, n), n, s, α) is the position in some play where II used τ . Note that α depends only on n, so we can denote it by αn . We show by induction on subformulas (φ, n) of φ that ( (φ, n), X n , dn ) ∈ T , where dn = 1 if and only if αn = II. Putting n = 1, we obtain α1 = II and we get the desired result. (i) Suppose (φ, n) is t1 = t2 or R(t1 , . . . , tn ). We show that the quadruple ( (φ, n), n, X n , d) is in T . Let s ∈ X n . Let the quadruple ( (φ, n), n, s, αn ) be a position in some play where II used τ . Since II wins with τ , ( (φ, n), X n , d) ∈ T . (ii) Suppose (φ, n) is =(t1 , . . . , tn ). Suppose first αn = II. Suppose s and s  are in X n and agree about the values of t1 , . . . , tn−1 . By the definition of

84

Game theoretic semantics

X n , ( (φ, n), n, s, II) and ( (φ, n), n, s  , II) are positions in some plays where II used τ . Since τ is uniform, s and s  agree about the value of tn . The case αn = I cannot occur since τ is a winning strategy. (iii) Suppose (φ, n) is ¬ψ. Note that X n = X n+1 . By the induction hypothesis, (ψ, X n , 1 − d) ∈ T , hence (¬ψ, X n , d) ∈ T . (iv) Suppose (φ, n) is (ψ ∨ θ ). Suppose first αn = I. Then both ( (φ, n + 1), n + 1, s, I) and ( (φ, n + 2 + len(ψ)), n + 2 + len(ψ), s, I) can be positions in some plays where II has used τ . By the induction hypothesis, (ψ, X n+1 , 0) ∈ T and (θ, X n+2+len(ψ) , 0) ∈ T . Note that X n ⊆ X n+1 ∩ X n+2+len(ψ) . Hence (ψ ∨ θ, X n , 0) ∈ T . Suppose then αn = II. Now X = Y ∪ Z , where Y is the set of s ∈ X n such that ( (φ, n + 1), n + 1, s, αn ) and Z is the set of s ∈ X n such that ( (φ, n + 2 + len(ψ)), n + 2 + len(ψ), s, αn ), By the induction hypothesis, (ψ, X n+1 , 1) ∈ T and (θ, X n+2+len(ψ) , 1) ∈ T . Hence (ψ ∧ θ, X n , 1) ∈ T . (v) Suppose (φ, n) is ∃xψ. We leave this as an exercise.  Exercise 5.24 Draw the game tree for H (φ), when φ is given by (i) (¬∃x0 =(x0 ) ∨ ¬∃x0 Rx0 ), (ii) ∃x0 ∃x1 (=(x1 ) ∨ Rx0 ). Exercise 5.25 Draw the game tree for H (φ), when φ is given by (i) ∀x0 ∃x1 (=(x1 ) ∨ Rx0 ), (ii) ∀x0 ∃x1 (P x0 ∨ ∃x2 (=(x0 , x2 ) ∨ Rx2 x1 )). Exercise 5.26 Draw the game tree for H (φ), when φ is given by ∃x0 (¬P x0 → ∀x1 (=(x0 , x1 ) ∧ Rx1 )). Exercise 5.27 Let L consist of two unary predicates P and R. Let M be an L-structure such that M = {0, 1, 2, 3}, P M = {0, 1, 2}, and R M = {1, 2, 3}. Who has a winning strategy in H (φ) if φ is given by the following: (i) ∃x0 (=(x0 ) ∧ Rx0 ), (ii) ∀x0 ∃x1 (=(x1 ) ∧ ¬(x0 = x1 ))? Describe the winning strategy. Exercise 5.28 Let M be as in Exercise 5.27. Does II have a uniform winning strategy in H (φ) if φ is given by the following: ∀x0 (P x0 → ∃x1 (=(x0 , x1 ) ∧ ¬(x0 = x1 ) ∧ P x0 ∧ Rx1 ))?

5.3 Imperfect information game for dependence logic

85

Table 5.2. x0

x1

∨

x0

x1

∨

0 1 2

0 1 2

left left right

0 1 2

0 1 2

left right right

Table 5.3. x0

x1

∨

x0

x1

∨

0 1 2

2 2 0

left

0 1 2

1 2 0

left

right

right

Exercise 5.29 Let M be as in Exercise 5.27. Does II have a uniform winning strategy in H (φ) if φ is given by following: ∃x0 (P x0 ∧ ∀x1 (=(x0 , x1 ) ∨ (x0 = x1 ) ∨ Rx1 ))? Exercise 5.30 Suppose M is the binary structure ({0, 1, 2}, R), where R is as in Fig. 5.6. Does II have a uniform winning strategy in H (φ) if φ is given by the following: ∀x0 ∃x1 (¬Rx0 x1 ∧ ∀x2 ∃x3 (=(x2 , x3 ) ∧ ¬Rx0 x3 ))? Exercise 5.31 Suppose M is the binary structure ({0, 1, 2}, R), where R is as in Fig. 5.6. Does II have a uniform winning strategy in H (φ) if φ is given by the following: ∃x0 ∀x1 (Rx0 x1 ∨ ∃x2 ∀x3 (=(x2 , x3 ) → Rx2 x3 ))? Exercise 5.32 Show that neither of the winning strategies (shown in Table 5.2) of player II in H (φ) is uniform, when φ is the sentence ∀x0 ∃x1 ((=(x1 ) ∧ x0 = x1 ) ∨ (=(x1 ) ∧ x0 = x1 )) and the universe is {0, 1, 2}. Exercise 5.33 Which of the strategies (in Table 5.3) of player II in H (φ) can be completed so that the strategy becomes a uniform winning strategy of II? Here φ is the sentence ∀x0 ∃x1 ((=(x1 ) ∧ ¬x0 = x1 ) ∨ (=(x1 ) ∧ ¬x0 = x1 ))) and the universe is {0, 1, 2}.

6 Model theory

Many model theoretic results for dependence logic can be proved by means of a reduction to existential second order logic. We establish this reduction in the first section of this chapter. This immediately gives such results as the Compactness Theorem, the L¨owenheim–Skolem Theorem, and the Craig Interpolation Theorem.

6.1 From D to 11 We associate with every formula φ of dependence logic a second order sentence which is in a sense equivalent to φ. This is in fact nothing more than a formalization of the truth definition of φ (Definition 3.5). What is interesting is that the second order sentence, which we denote by τ1,φ (S), is not just any second order sentence but a particularly simple second order existential sentence, called a 11 -sentence. Such sentences have a close relationship with first order logic, especially on countable models. It turns out that their relationship with dependence logic is even closer. In a sense they are one and the same thing. It is the main purpose of this section to explain exactly what is this sense in which they are one and the same thing. Definition 6.1 Let L be a vocabulary. The class of 11 -formulas of L is defined as follows. (i) Any first order formula of L is a 11 -formula of L. (ii) If φ is a 11 -formula of L ∪ {R}, then ∃Rφ is a 11 -formula of L. M |=s ∃Rφ if and only if there is an expansion M of M to an L ∪ {R}-structure such that M |=s φ. 86

6.1 From D to 11

87

(iii) If φ is a 11 -formula of L ∪ { f }, then ∃ f φ is a 11 -formula of L. M |=s ∃ f φ if and only if there is an expansion M of M to an L ∪ { f }-structure such that M |=s φ. An equivalent concept is PC-definability. A property O(M) of L-structures is PC-definable if there is a vocabulary L  ⊇ L and a first order φ in the vocabulary L  such that an L-structure M has the property O(M) if and only if it is a reduct of an L  -structure satisfying φ. Clearly, PC-definable properties are exactly the 11 -definable properties of models. The logic 11 is closed under conjunction in the sense that if φ and ψ are 1 1 -formulas then there is a 11 -formula θ such that M |=s θ ⇐⇒ M |=s φ and M |=s ψ. For example, if φ = ∃R∃R  φ  and ψ = ∃R∃ f ψ  , then we first change the relation symbol R in ψ  to a new relation symbol R  , obtaining ψ  , and then take ∃R∃R  ∃R  ∃ f (φ  ∧ ψ  ) as θ . Similarly, 11 is closed under disjunction in the sense that if φ and ψ are 11 -formulas then there is a 11 -formula θ such that M |=s θ ⇐⇒ M |=s φ or M |=s ψ. It is fairly obvious that 11 is closed under first order existential quantification in the sense that if φ is an 11 -formula then there is a 11 -formula θ such that M |=s θ ⇐⇒ there is a ∈ M such that M |=s(a/xn ) φ. For example, if φ = ∃R∃R  ψ, then we can take ∃R∃R  ∃xn ψ as θ. It is a little more tricky to see that 11 is also closed under first order universal quantification in the sense that if φ is an 11 -formula then there is a 11 -formula θ such that M |=s θ ⇐⇒ for all a ∈ M we have M |=s(a/xn ) φ. To see why this is so, let us consider a simple 11 -formula ∃Rψ, where ψ is first order. Let φ  be obtained from φ by replacing everywhere Rt1 . . . tm by R  xn t1 . . . tn , where R  is a new predicate symbol of arity # L (R) + 1. Now, M |=s ∃R  ∀xn φ  ⇐⇒ for all a ∈ M we have M |=s(a/xn ) ∃Rφ

(6.1)

(see Exercise 6.7). We adopt the convention of writing a formula φ of dependence logic with free variables xi1 , . . . , xin as φ(xi1 , . . . , xin ), where it is always assumed that i 1 < · · · < i n . This notation includes the case that n = 0 which corresponds to

Model theory

88

the case that φ is a sentence. Similarly we write t(xi1 , . . . , xin ) for a term built up from the variables xi1 , . . . , xin . In the following theorem we refer to S as an n-ary predicate symbol. If n = 0, S is a  or ¬. In the following theorem, rel(X ) = {(s(xi1 ), . . . , s(xin )) : s ∈ X }. Theorem 6.2 We can associate with every formula φ(xi1 , . . . , xin ) of D in vocabulary L and every d ∈ {0, 1} a 11 -sentence τd,φ (S), where S is n-ary, such that for all L-structures M and teams X with dom(X ) = {xi1 , . . . , xin } the following are equivalent: (i) (φ, X, d) ∈ T , (ii) (M, rel(X )) |= τd,φ (S). Proof We modify Hodges’ approach (see ref. [22], sect. 3) to fit our setup. The sentence τd,φ (S) is simply Definition 3.5 written in another way. There is nothing new in τd,φ (S), and in each case the proof of the claimed equivalence is straightforward (see Exercises 6.4 and 6.5). Case (1) Suppose φ(xi1 , . . . , xin ) is t1 = t2 or Rt1 . . . tn . We rewrite (D1), (D2), (D5) and (D6) of Definition 3.5 by letting τ1,φ (S) be ∀xi1 . . . ∀xin (Sxi1 . . . xin → φ(xi1 , . . . , xin )) and by letting τ0,φ (S) be ∀xi1 . . . ∀xin (Sxi1 . . . xin → ¬φ(xi1 , . . . , xin )). Case (2) Suppose φ(xi1 , . . . , xin ) is the dependence formula =(t1 (xi1 , . . . , xin ), . . . , tm (xi1 , . . . , xin )), where i 1 < · · · < i n . Recall conditions (D3) and (D4) of Definition 3.5. Following these conditions, we define τ1,φ (S) as follows. Subcase (2.1) m = 0. We let τ1,φ (S) =  and τ0,φ (S) = ¬. Subcase (2.2) m = 1. Now φ(xi1 , . . . , xin ) is the dependence formula = (t1 (xi1 , . . . , xin )). We let τ1,φ (S) be the formula ∀xi1 . . . ∀xin ∀xin +1 . . . ∀xin +n ((Sxi1 . . . xin ∧ Sxin +1 . . . xin +n ) → t1 (xi1 , . . . , xin ) = t1 (xin +1 , . . . , xin +n )) and we further let τ0,φ (S) be the formula ∀xi1 . . . ∀xin ¬ Sxi1 . . . xin .

6.1 From D to 11

89

Subcase (2.3) If m > 1 we let τ1,φ (S) be the formula ∀xi1 . . . ∀xin ∀xin +1 . . . ∀xin +n ((Sxi1 . . . xin ∧ Sxin +1 . . . xin +n ∧ t1 (xi1 , . . . , xin ) = t1 (xin +1 , . . . , xin +n ) ∧ ... tm−1 (xi1 , . . . , xin ) = tm−1 (xin +1 , . . . , xin +n )) → tm (xi1 , . . . , xin ) = tm (xin +1 , . . . , xin +n )), and we further let τ0,φ (S) be the formula ∀xi1 . . . ∀xin ¬ Sxi1 . . . xin . Case (3) Suppose φ(xi1 , . . . , xin ) is the disjunction (ψ(x j1 , . . . , x j p ) ∨ θ (xk1 , . . . , xkq )), where {i 1 , . . . , i n } = { j1 , . . . , j p } ∪ {k1 , . . . , kq }. We let the sentence τ1,φ (S) be ∃R∃T (τ1,ψ (R) ∧ τ1,θ (T )∧ ∀xi1 . . . ∀xin (Sxi1 . . . xin → (Rx j1 . . . x j p ∨ T xk1 . . . xkq ))) and we let the sentence τ0,φ (S) be ∃R∃T (τ0,ψ (R) ∧ τ0,θ (T )∧ ∀xi1 . . . ∀xin (Sxi1 . . . xin → (Rx j1 . . . x j p ∧ T xk1 . . . xkq ))). Case (4) φ is ¬ψ. τd,φ (S) is the formula τ1−d,ψ (S). Case (5) Suppose φ(xi1 , . . . , xin ) is the formula ∃xin+1 ψ(xi1 , . . . , xin+1 ). τ1,φ (S) is the formula ∃R(τ1,ψ (R) ∧ ∀xi1 . . . ∀xin (Sxi1 . . . xin → ∃xin+1 Rxi1 . . . xin+1 )) and τ0,φ (S) is the formula ∃R(τ0,ψ (R) ∧ ∀xi1 . . . ∀xin (Sxi1 . . . xin → ∀xin+1 Rxi1 . . . xin+1 )). 

Corollary 6.3 For every sentence φ of D there are 11 -sentences τ1,φ and τ0,φ such that for all models M we have M |= φ if and only if M |= τ1,φ ; M |= ¬φ if and only if M |= τ0,φ . Proof Let τd,φ be the result of replacing in τd,φ (S) every occurrence of the 0-ary relation symbol S by . Now the claim follows from Theorem 6.2.  Exercise 6.1 Write down τ1,∃x1 =(x1 ) (S) and τ0,∃x1 =(x1 ) (S). Exercise 6.2 What is τ1,φ (S) if φ is ∃x1 (=(x1 ) ∨ x1 = x0 )?

90

Model theory

Exercise 6.3 What is τ1,φ (S) if φ is given by the following: ∀x0 ∃x1 ∀x2 ∃x3 (=(x2 , x3 ) ∧ ¬(x1 = x4 ) ∧ (x0 = x2 ↔ x1 = x3 ))? Exercise 6.4 Fill in the details of Case (3) of the proof of Theorem 6.2. Exercise 6.5 Fill in the details of Case (5) of the proof of Theorem 6.2. Exercise 6.6 Show that if φ is 11 , M |= φ and M ∼ = N , then N |= φ. Exercise 6.7 Prove the claim given in Eq. (6.1).

6.2 Applications of 11 The 11 -representation of D-formulas yields some immediate but nevertheless very important applications. They are all based on model theoretic properties of first order logic, which we now review. Compactness Theorem of first order logic Suppose T is an arbitrary set of sentences of first order logic such that every finite subset of T has a model. Then T itself has a model. There are many different proofs of this classical result due to G¨odel and Mal’cev. The proof is somewhat easier in the case that T is countable. One main line of proof [15] constructs a sufficiently complete extension T ∗ of T in a vocabulary which has infinitely many new constant symbols. After this a syntactical (term-) model is constructed by means of T ∗ . In another line of proof a model of T is constructed by gluing together the models of the finite parts of T into one structure which models all of T . This is the ultraproduct approach [9]. L¨owenheim–Skolem Theorem of first order logic Suppose φ is a sentence of first order logic such that φ has an infinite model or arbitrarily large finite models. Then φ has models of all infinite cardinalities [36], [37], [39]. This is one of the oldest results of model theory of first order logic. The proof combines the method of Skolem functions and compactness. Craig Interpolation Theorem of first order logic Suppose φ and ψ are sentences of first order logic such that |= φ → ψ. Suppose the vocabulary of φ is L φ and that of ψ is L ψ . Then there is a first order sentence θ of vocabulary L φ ∩ L ψ such that |= φ → θ and |= θ → ψ. There are several different proofs of this result. The original proof of William Craig [6] was proof

6.2 Applications of 11

91

theoretic, based on cut-elimination. Most model theoretic proofs, starting with ref. [34], use compactness in one form or another. We can now easily derive similar results for dependence logic by appealing to the 11 -representation of D-sentences. Theorem 6.4 (Compactness Theorem of D) Suppose  is an arbitrary set of sentences of dependence logic such that every finite subset of  has a model. Then  itself has a model. Proof Let  = {φi : i ∈ I } and let L be the vocabulary of . Let τ1,φi = ∃S1i . . . ∃Sni i ψi , where ψi is first order. By changing symbols we can assume that all S ij are different symbols. Let T be the first order theory {ψi : i ∈ I } in the vocabulary L  = L ∪ {S ij : i ∈ I, 1 ≤ j ≤ n i }. Every finite subset of T has a model. By the Compactness Theorem of first order logic, there is an L  structure M that is a model of the theory T itself. The reduction M = M L of M to the original vocabulary L is, by definition, a model of .  Theorem 6.5 (L¨owenheim–Skolem Theorem of D) Suppose φ is a sentence of dependence logic, such that φ either has an infinite model or has arbitrarily large finite models. Then φ has models of all infinite cardinalities, in particular φ has a countable model and an uncountable model. Proof Let τ1,φ = ∃S1 . . . ∃Sn ψ, where ψ is first order in the vocabulary L  = L ∪ {S1 , . . . , Sn }. Suppose κ is an arbitrary infinite cardinal number. By the L¨owenheim–Skolem Theorem of first order logic, there is an L  -model M of ψ of cardinality κ. The reduction M = M L of M to the original vocabulary L is a model of φ of cardinality κ.  Corollary 6.6 (ref. [31]) A sentence of dependence logic in the empty vocabulary is true in one infinite model (or arbitrarily large finite ones) if and only it is true in every infinite model. Proof All models of the empty vocabulary of the same cardinality are isomorphic. Thus the claim follows from Theorem 6.5.  We shall address the Craig Interpolation Theorem in Corollary 6.17 and derive first a Separation Theorem which is an equivalent formulation in the case of first order logic. Theorem 6.7 (Separation Theorem) Suppose φ and ψ are sentences of dependence logic such that φ and ψ have no models in common. Let the vocabulary of φ be L and the vocabulary of ψ be L  . Then there is a sentence θ of D in the vocabulary L ∩ L  such that every model of φ is a model of θ, but θ and ψ have no models in common. In fact, θ can be chosen to be first order.

92

Model theory

Proof Let τ1,φ = ∃S1 . . . ∃Sn φ0 , where φ0 is first order in the vocabulary L 0 . Let τ1,ψ = ∃S1 . . . ∃Sm ψ0 , where ψ0 is first order in the vocabulary L 0 . Without loss of generality, {S1 , . . . , Sn } ∩ {S1 , . . . , Sm } = ∅. Note that |= φ0 → ¬ψ0 for if M is a model of φ0 ∧ ψ0 , then ML |= φ ∧ ψ, contrary to the assumption that φ and ψ have no models in common. By the Craig Interpolation Theorem for first order logic, there is a first order sentence θ of vocabulary L ∩ L  such that |= φ0 → θ and |= θ → ¬ψ0 . Every model of φ is a model of θ , but θ and ψ have no models in common.  A particularly striking application of Theorem 6.7 is the following special case in which φ and ψ not only have no models in common but, furthermore, every model satisfies one of them. Theorem 6.8 (Failure of the Law of Excluded Middle) Suppose φ and ψ are sentences of dependence logic such that for all models M we have M |= φ if and only if M |= ψ. Then φ is logically equivalent to a first order sentence θ such that ψ is logically equivalent to ¬θ . Proof The first order θ obtained in the proof of Theorem 6.7 is the θ we seek.  Note that it is perfectly possible to have for all finite models M M |= φ if and only if M |= ψ without φ or ψ being logically equivalent to a first order sentence. For example, in the empty vocabulary φ can say the size of the universe is even while ψ says it is odd. Definition 6.9 A sentence φ of dependence logic is called determined in M if M |= φ or M |= ¬φ. Otherwise φ is called non-determined in M. We say that φ is determined if φ is determined in every structure. A typical non-determined sentence (from ref. [2]) is given by ∀x0 ∃x1 (=(x1 ) ∧ x0 = x1 ), which is non-determined in every structure with at least two elements. The following corollary shows that it is not at all difficult to find other non-determined sentences. Corollary 6.10 Every determined sentence of dependence logic is strongly logically equivalent to a first order sentence.

6.2 Applications of 11

93

Proof Suppose φ is determined. Thus for all M we have M |= φ or M |= ¬φ. It follows that for all M we have M |= φ if and only if M |= ¬φ. By Theorem 6.8 there is a first order θ such that φ is logically equivalent to θ and ¬φ is logically equivalent to ¬θ . Thus φ is strongly logically equivalent to θ.  Thus we can take any sentence of dependence logic, which is not strongly equivalent to a first order sentence, and we know that there are models in which the sentence is non-determined. Example 6.11 The sentence wf is non-determined in every infinite well ordered structure. This can be seen either by a direct argument based on the truth definition, or by the following indirect argument. Suppose M is an infinite well ordered linear order in which wf is determined. Thus M |= ¬ wf . Let  be the following set of sentences of dependence logic: ¬ wf , c1 < c0 , c2 < c1 , ··· cn+1 < cn , ··· It is evident that every finite subset of  is true in an expansion of M. By the Compactness Theorem, there is a model M of the whole . Then M is ill-founded and therefore satisfies wf . This contradicts the fact that M also satisfies ¬ wf . For more examples of non-determinacy, see ref. [42]. Exercise 6.8 Show that the sentence wf is non-determined in all sufficiently big finite linear orders. Exercise 6.9 Show that even is non-determined in every sufficiently large finite model of odd size. Exercise 6.10 Show that ∞ is non-determined in every sufficiently big finite model. Exercise 6.11 Show that cmpl is non-determined in every complete dense linear order. Exercise 6.12 Suppose φn , n ∈ N, are sentences of D such that each φn is true in some model, and moreover φn+1 ⇒ φn for all n. Show that there is one model M in which each φn is true.

Model theory

94

Exercise 6.13 Show that if φ is a sentence of D and ψ is a first order sentence such that every countable1 model of φ is a model of ψ, then every model of φ is a model of ψ. Exercise 6.14 Give a sentence φ of D and a first order sentence ψ such that every countable model of ψ is a model of φ and vice versa, but some model of ψ is not a model of φ.

6.3 From 11 to D We have seen that representing formulas of dependence logic in 11 form is a powerful method for getting model theoretic results about dependence logic. We now show that this method is in a sense the best possible. Namely, we can also translate any 11 -sentence back to dependence logic. We prove first a fundamental property of first order and 11 -formulas. Its various formulations all carry the name of Thoralf Skolem [36] (see also ref. [37]), who proved the following result in 1920. The basic idea is that the existential second order quantifiers in front of 11 -formulas are so powerful that they subsume all other existential quantifiers. Theorem 6.12 (Skolem Normal Form Theorem) Every 11 -formula φ is logically equivalent to an existential second order formula ∃ f 1 . . . ∃ f n ∀x1 . . . ∀xm ψ,

(6.2)

where ψ is quantifier free and f 1 , . . . , f n are function symbols. The formula in Eq. (6.2) is called a Skolem Normal Form of φ. Proof For atomic and negated atomic φ we choose φ itself as a Skolem Normal Form. Suppose then φ0 has a Skolem Normal Form, ∃ f 10 . . . ∃ f n00 ∀x10 . . . ∀xm0 0 ψ0 ,

(6.3)

and φ1 has a Skolem Normal Form, ∃ f 11 . . . ∃ f n11 ∀x11 . . . ∀xm1 1 ψ1 .

(6.4)

By changing bound variables, we may assume, w.l.o.g., that { f 10 , . . . , f n00 } ∩ { f 11 , . . . , f n11 } = ∅ and that {x10 , . . . , xm0 0 } ∩ {x11 , . . . , xm1 1 } = ∅. Now we obtain a Skolem Normal Form, ∃ f 10 . . . ∃ f n00 ∃ f 11 . . . ∃ f n11 ∀x10 . . . ∀xm0 0 ∀x11 . . . ∀xm1 1 (ψ0 ∧ ψ1 ), 1

i.e. countably infinite or finite.

6.3 From 11 to D

95

for φ0 ∧ φ1 and a Skolem Normal Form, ∃ f 10 . . . ∃ f n00 ∃ f 11 . . . ∃ f n11 ∀x10 . . . ∀xm0 0 ∀x11 . . . ∀xm1 1 (ψ0 ∨ ψ1 ), for φ0 ∨ φ1 . For a definition of a Skolem Normal Form for ∀x0 φ, suppose φ has a Skolem Normal Form as given in Eq. (6.2). Let f 1 , . . . , f n be new function symbols such that #( f i ) = #( f i ) + 1 and let x0 be a variable not occurring in ψ so that also x0 ∈ / {x1 , . . . , xn }. Let ψ  be obtained from ψ by inductively replacing everywhere f i (t1 , . . . , t#( fi ) ) by f i (x0 , t1 , . . . , t#( fi ) ). We define ∃ f 1 . . . ∃ f n ∀x0 ∀x1 . . . ∀xm ψ  to be the Skolem Normal Form of ∀x0 φ. Let f 0 be a new 0-place function variable and let ψ  be obtained from ψ by replacing everywhere x0 by f 0 . We let ∃ f 0 ∃ f 1 . . . ∃ f n ∀x1 . . . ∀xm ψ  be a Skolem Normal Form of ∃x0 φ. For the definition of a Skolem Normal Form for ∃Rφ, suppose φ has the Skolem Normal Form given in Eq. (6.2). A Skolem Normal Form of ∃Rφ is simply ∃ f 1 . . . ∃ f n ∃ f n+1 ∃ f n+2 ∀x1 . . . ∀xm θ, where θ is obtained from ψ by replacing everywhere Rt1 . . . tk by the formula f n+1 t1 . . . tk = f n+2 . Finally, the construction of a Skolem Normal Form of ∃ f φ is easy. Suppose φ has the Skolem Normal Form as in Eq. (6.2). We obtain the Skolem Normal Form ∃ f ∃ f 1 . . . ∃ f n ∀x1 . . . ∀xm ψ for ∃ f φ.



Example 6.13 The sentence ∀x1 ∃x2 Rx1 x2 in Skolem Normal Form reads as follows: ∃ f ∀x1 Rx f x1 , and ∀x1 (∃x2 P x1 x2 ∨ ∀x3 ∃x4 Rx1 x3 x4 ) reads as follows: ∃ f 1 ∃ f 2 ∀x1 ∀x3 (P x1 f 1 x1 ∨ Rx1 x3 f 2 x1 x3 ).

Model theory

96

Corollary 6.14 (Skolem Normal Form Theorem for D) For every formula ∗ φ of dependence logic and every d ∈ {0, 1}, there is a 11 -sentence τd,φ (S) of the following form: ∃ f 1 . . . ∃ f n ∀x1 . . . ∀xm ψ,

(6.5)

where ψ is quantifier-free, such that the following are equivalent: (i) (φ, X, d) ∈ T , ∗ (ii) (M, X ) |= τd,φ (S). In particular, for every sentence φ of dependence logic there are 11 -sentences ∗ ∗ τ1,φ and τ0,φ of the form (6.5) such that for all models M we have ∗ M |= φ if and only if M |= τ1,φ . ∗ M |= ¬φ if and only if M |= τ0,φ .

Corollary 6.14 gives an easy proof of the L¨owenheim–Skolem Theorem for dependence logic (Theorem 6.5). Namely, suppose φ is a given sentence of dependence logic with an infinite model or arbitrarily large finite models. ∗ By compactness we may assume φ indeed has an infinite model M. Let τ1,φ  be of the form ∃ f 1 . . . ∃ f n ∀x1 . . . ∀xm ψ. Thus there are interpretations f iM of the function symbols f i in an expansion of M of M such that M satisfies ∀x1 . . . ∀xm ψ. Let N be a countable subset of M such that N is closed under  all the n functions f iM . Because ∀x1 . . . ∀xm ψ is universal, it is still true in the countable substructure M∗ of M generated by N . Thus M∗ is a countable model of φ. This is in line with the original proof of Skolem. If we wanted a model of size κ for a given infinite cardinal number κ, the argument would be similar, but we would first use compactness to get a model of size at least κ. Theorem 6.15 (refs. [7] and [44]) For every 11 -sentence φ there is a sentence φ ∗ in dependence logic such that for all M: M |= φ ⇐⇒ M |= φ ∗ . Proof We may assume φ is of the following form: = ∃ f 1 . . . ∃ f n ∀x1 . . . ∀xm ψ,

(6.6)

where ψ is quantifier-free. We will perform some reductions on Eq. (6.6) in order to make it more suitable for the construction of φ. Step (1) If ψ contains nesting of the function symbols f 1 , . . . , f n or of the function symbols of the vocabulary, we can remove them one by one

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97

by using the equivalence of |= ψ( f i t1 . . . tm ) and ∀x1 . . . ∀xm ((t1 = x1 ∧ . . . ∧ tm = xm ) → ψ( f i x1 . . . xm )). Thus we may assume that all terms occurring in ψ are of the form xi or f i xi1 . . . xik . Step (2) If ψ contains an occurrence of a function symbol f i xi1 . . . xik with the same variable occurring twice, e.g. i s = ir , 1 < r < k, we can remove it by means of a new variable xl and the equivalence |= ∀x1 . . . ∀xm ψ( f i xi1 . . . xik ) ↔ ∀x1 . . . ∀xm ∀xl (xl = xr → ψ( f i xi1 . . . xir −1 , xl , xir +1 . . . xik )). Thus we may assume that if a term such as f i xi1 . . . xik occurs in ψ, its variables are all distinct. Step (3) If ψ contains two occurrences of the same function symbol but with different variables or with the same variables in different order, we can remove it using appropriate equivalences. If {i 1 , . . . , i k } ∩ { j1 , . . . , jk } = ∅, we have the equivalence |= ∀x1 . . . ∀xm ψ( f i xi1 . . . xik , f i x j1 . . . x jk ) ↔ ∃ f i ∀x1 . . . ∀xm (ψ( f i xi1 . . . xik , f i x j1 . . . x jk ) ∧ ((xi1 = x j1 ∧ . . . ∧ xik = x jk ) → f i xi1 . . . xik = f i x j1 . . . x jk )). We can reduce the more general case, where {i 1 , . . . , i k } ∩ { j1 , . . . , jk } = ∅, to this case by introducing new variables, as in Step (2). (We are grateful to Ville Nurmi for pointing out the necessity of this.) Thus we may assume that for each function symbol f i occurring in ψ, there are j1i , . . . , jni i such that all occurrences of f i are of the form f i x j1i . . . x jmi and j1i , . . . , jmi i are all different i from each other. In sum, we may assume the function terms that occur in ψ are of the form f i x j1i . . . x jmi and for each i the variables x j1i , . . . , x jmi and their order is the i i same. Let N be greater than all the x jki . Following the notation of Eq. (6.6), let φ ∗ be the sentence ∀x1 . . . ∀xm ∃x N +1 . . . ∃x N +n (=(x j11 , . . . , x jm1 , x N +1 ) ∧ 1

... (=(x j1n , . . . , x jmn n , x N +n ) ∧ ψ  ),

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where ψ  is obtained from ψ by replacing everywhere f i x j1i . . . x jmi by x N +i . i This is clearly the desired sentence. 

It is noteworthy that the D-representation of a given 11 -sentence given above is of universal-existential form, i.e. of the form ∀xn 1 . . . ∀xn k ∃xm 1 . . . ∃xm l ψ, where ψ is quantifier-free. Moreover, ψ is just a conjunction of dependence statements =(x1 , . . . , xn ) and a quantifier-free first order formula. This is a powerful normal form for dependence logic. Corollary 6.16 For any sentence φ of dependence logic of vocabulary L and for every L  ⊆ L there is a sentence φ  of dependence logic of vocabulary L  such that the following are equivalent: (i) M |= φ  ; (ii) there is an expansion N of M such that N |= φ. Proof Let ψ be a 11 -sentence logically equivalent with φ. We assume for simplicity that L \ L  consists of just one predicate symbol R. Let φ  be a sentence of dependence logic logically equivalent with the 11 -sentence ∃Rψ. Then φ  is a sentence satisfying the equivalence of the conditions (i) and (ii).  Corollary 6.16 implies in a trivial way the following strong form of the Craig Interpolation Theorem. Corollary 6.17 (uniform interpolation property) Suppose φ is a sentence of D. Let L be the vocabulary of φ. For every L  ⊆ L there is a sentence φ  of D in the vocabulary L  which is a uniform interpolant of φ in the following sense: φ ⇒ φ  , and, if ψ is a sentence of D in a vocabulary L  such that φ ⇒ ψ and L ∩ L  = L  , then φ  ⇒ ψ. Proof Let φ  be as in Corollary 6.16. By its very definition, φ  is a logical consequence of φ. Suppose then ψ is a sentence of D in a vocabulary L  such that φ ⇒ ψ and L ∩ L  = L  . If M is an L  -structure which is a model of φ  , then M L  is a model of φ  , whence there is an expansion M of M L  to a model of φ. Since φ ⇒ ψ, M is a model of ψ. But ML  = M L  . Thus M |= ψ.  For a version of the Beth Definability Theorem, see Exercise 6.23.

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Exercise 6.15 Give a Skolem Normal Form for the following first order formulas: (i) ∀x0 ∃x1 x0 = x1 ; (ii) ∃x0 ∀x1 ¬x0 = x1 ; (iii) (∃x0 P x0 ∨ ∀x0 ¬P x0 ). Exercise 6.16 Give a Skolem Normal Form for the following first order formula: ∀x0 ∃x1 ∀x2 ∃x3 ((x0 = x4 → ¬(x1 = x4 )) ∧ (x0 < x3 ↔ x1 < x2 )). Exercise 6.17 Write the following 11 -sentences in Skolem Normal Form: (i) ∃x0 ∃ f ∀x1 (¬ f x1 = x0 ∧ ∀x2 ( f x0 = f x1 → x0 = x1 )); (ii) ∃R(∀x0 ∀x1 ∀x2 ((Rx0 x1 ∧ Rx1 x2 ) → Rx0 x2 ) ∧ ∀x0 ∀x1 (Rx0 x1 ∨ Rx1 x0 ∨ x0 = x1 ) ∧ ∀x0 ¬Rx0 x0 ∧ ∀x0 ∃x1 Rx0 x1 ). Exercise 6.18 Give a sentence of D which is logically equivalent to the following 11 -sentence in Skolem Normal Form: (i) ∃ f 0 ∃ f 1 ∀x0 ∀x1 φ(x0 , x1 , f 0 (x0 , x1 ), f 1 (x0 , x1 )); (ii) ∃ f 0 ∃ f 1 ∀x0 ∀x1 φ(x0 , x1 , f 0 (x0 , x1 ), f 1 (x1 )); (iii) ∃ f 0 ∃ f 1 ∀x0 φ(x0 , f 0 (x0 ), f 1 (x0 )). In each case φ is quantifier-free and first order. Exercise 6.19 Give a sentence of D which is logically equivalent to the following 11 -sentence: ∃ f ∀x0 ∀x1 φ(x0 , x1 , f (x0 , x1 ), f (x1 , x0 )), where φ is quantifier-free. Exercise 6.20 Express the Henkin quantifier   ∀x ∃y R(x, y, u, v) ↔ ∃ f ∃g∀x∀u R(x, f (x), u, g(u)) ∀u ∃v in dependence logic. Exercise 6.21 Suppose M is an L-structure and P ⊆ M n . We say that P is D-definable in M if there is a sentence φ(c1 , . . . , cn ) of D with new constant symbols c1 , . . . , cn such that the following are equivalent for all a1 , . . . , an ∈ M: (i) (a1 , . . . , an ) ∈ P;

Model theory

100 (ii) (M, a1 , . . . , an ) |= φ,

where (M, a1 , . . . , an ) denotes the expansion of M obtained by interpreting ci in (M, a1 , . . . , an ) by ai . We then say that φ defines P in M. Show that if P and Q are D-definable, then so are P ∩ Q and P ∪ Q, but not necessarily P \ Q. Exercise 6.22 Recall the definition of D-definability in a model in Exercise 6.21. Let L be a vocabulary. Suppose ψ is a D-sentence in a vocabulary L ∪ {R}, where R is a new n-ary predicate symbol. We say that R is D-definable in models of ψ if there is a D-sentence φ of vocabulary L ∪ {c1 , . . . , cn } such that φ defines R in every model of ψ. Prove the following useful criterion for D-undefinability: if ψ has two models M and N such that ML = N L but R M = R N , then R is not D-definable in models of ψ. (In the case of first order logic this is known as the Padoa Principle.) Exercise 6.23 Recall the definition of D-definability in models of a sentence in Exercise 6.22. Let L be a vocabulary. Suppose ψ is a D-sentence in a vocabulary L ∪ {R}, where R is a new n-ary predicate symbol. Suppose any two models M and N of ψ such that ML = N L satisfy also R M = R N . Show that R is D-definable in models of ψ (In the case of first order logic this is known as the Beth Definability Theorem.) Exercise 6.24 (See ref. [3].) Suppose φ and ψ are sentences of D such that no model satisfies both φ and ψ. Show that there is a sentence θ of D such that M |= φ if and only if M |= θ and M |= ψ if and only if M |= ¬θ.

6.4 Truth definitions In 1933, the Polish logician Alfred Tarski defined the concept of truth in a general setting (see, e.g., ref. [38]) and pointed out what is known as Tarski’s Undefinability of Truth argument: no language can define its own truth, owing to the Liar Paradox, namely to the sentence “This sentence is false.” This sentence is neither true nor false, contrary to the Law of Excluded Middle, which Tarski took for granted. By 1931, the Austrian logician Kurt G¨odel [12],

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101

working not on arbitrary formalized languages but on first order number theory, had constructed, using a lengthy process referred to as the arithmetization of syntax, the following sentence: “This sentence is unprovable.” This sentence cannot be provable, for then it would be true, and hence unprovable. So it is unprovable and hence true. Its negation cannot be provable either, for else the negation would be true. So it is an example of a sentence which is independent of first order number theory. This is known as G¨odel’s First Incompleteness Theorem. G¨odel’s technique could be used to make exact sense of undefinability of truth (see below) and to prove it exactly for first order number theory. Exercise 6.25 Consider the following: “If this sentence is true, then its negation is true.” Derive a contradiction. Exercise 6.26 Consider the following: “It is not true that this sentence is true.” Derive a contradiction. Exercise 6.27 (See ref. [28].) Consider the following sentences. (1) It is raining in Warsaw. (2) It is raining in Vienna. (3) Exactly one of the sentences (1)–(3) is true. Under what kind of weather conditions in Europe is sentence (3) paradoxical?

6.4.1 Undefinability of truth Even to formulate the concept of definability of truth, we have to introduce a method for speaking about a formal language in the language itself. The clearest way of doing this is by means of G¨odel-numbering. Each sentence φ is associated with a natural number φ, its G¨odel-number, in a systematic way, described in Section 6.4.2. Moreover, we assume that our language has a name n for each natural number n. Definition 6.18 A truth definition for any model M and any formal language L, such as first order logic or dependence logic, is a formula τ (x0 ) of some possibly other formal language L such that for each sentence φ of L we have M |= φ if and only if M |= τ (φ).

(6.7)

A stronger requirement would be M |= φ ↔ τ (φ), but this would be true only in the presence of the Law of Excluded Middle, as (φ ↔ ψ) ⇒ (φ ∨ ¬φ).

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Model theory

An even stronger requirement would be the provability of φ ↔ τ (φ) from some axioms, but we abandon this also in the current setup. By the vocabulary L {+,×} of arithmetic we mean a vocabulary appropriate for the study of number theory, with a symbol N for the set of natural numbers. We specify L {+,×} in detail below. We call an L-structure Mω , where L ⊇ L {+,×} , “a model of Peano’s axioms” if the reduct of Mω to the vocabulary {N , +, ×} satisfies the first order Peano axioms of number theory. The results on definability of truth are relevant even if we assume that N Mω is the whole universe of the model Mω . Below, Mω denotes such a model of Peano’s axioms. Theorem 6.19 (G¨odel’s Fixed Point Theorem) For any first order formula φ(x0 ), in the vocabulary of arithmetic there is a first order sentence ψ of the same vocabulary such that for all models Mω of Peano’s axioms, Mω |= ψ if and only if Mω |= φ(ψ). Proof Let Sub be the set of triples w, w , n, where w  is obtained from w by replacing x0 by the term n. Since recursive relations are representable in models of Peano’s axioms, there is a first order formula σ (x0 , x1 , x2 ) such that n, m, k ∈ Sub ⇐⇒ Mω |= σ (n, m, k). W.l.o.g., x0 is not bound in σ (x0 , x1 , x2 ) and x0 and x1 are not bound in φ(x0 ). Let θ(x0 ) be the formula ∃x1 (φ(x1 ) ∧ σ (x0 , x1 , x0 )). Let k = θ (x0 ) and ψ = θ(k). Then Mω |= ψ if and only if Mω |= φ(ψ).  The above result does not hold just for first order logic but for any extension of first order logic, the syntax of which is sufficiently effectively given, for example dependence logic. Theorem 6.20 (Tarski’s Undefinability of Truth Result) First order logic does not have a truth definition in first order logic for any model Mω of Peano’s axioms. Proof Let τ (x0 ) be as in Definition 6.18. By Theorem 6.19 there is a sentence ψ such that Mω |= ψ if and only if Mω |= ¬τ (ψ).

(6.8)

If Mω |= ψ, then Mω |= τ (ψ) by Eq. (6.7), and Mω |= ¬τ (ψ) by Eq. (6.8). Hence Mω |= ψ. Now Mω |= τ (ψ) by Eq. (6.7), and Mω |= ¬τ (ψ) by Eq. (6.8). So neither τ (ψ) nor ¬τ (ψ) is true in Mω . This contradicts the Law of Excluded Middle, which says in this case Mω |= (τ (ψ) ∨ ¬τ (ψ)). 

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103

Theorem 6.20 has many stronger formulations. As the almost trivial proof above shows, no language for which the G¨odel Fixed Point Theorem can be proved and which satisfies the Law of Excluded Middle for its negation can have a truth definition in the language itself. We shall not elaborate more on this point here, as our topic, dependence logic, certainly does not satisfy the Law of Excluded Middle for its negation. Exercise 6.28 Prove Mω |= ψ if and only if Mω |= φ(ψ) in the proof of Theorem 6.19.

6.4.2 Definability of truth in first order logic We turn to another important contribution of Tarski, namely that truth is implicitly (or even better – inductively) definable in first order logic. In dependence logic the implicit definition can even be turned into an explicit definition by means of Theorem 6.15, as emphasized by Hintikka [19]. So, after all, truth is definable, albeit only implicitly. The realization of this may be an even more important contribution of Tarski to logic than the undefinability of truth. We shall carry out in some detail the definition of truth for first order logic. We shall omit many details, as these are well covered by the literature. For simplicity, we assume that the vocabulary L {+,×} of arithmetic includes all the machinery needed for arithmetization. All that we really need is a pairing function, but the pursuit of such minimalism is not relevant for the main argument and belongs to other contexts. Another simplification is that we only consider truth in models Mω of Peano’s axioms. We consider a finite vocabulary L = {c1 , . . . , cn , R1 , . . . , Rm , f 1 , . . . , f k } containing L {+,×} . When we specify in the following what L {+,×} should contain we assume they are all among c1 , . . . , cn , R1 , . . . , Rm , f 1 , . . . , f k . Let ri = #(Ri ). If w = w0 . . . wk is a string of symbols in the following alphabet: =, ci , Ri , f i , (, ), ¬, ∨, ∃, the G¨odel-number w of w is the natural number given by w = p0 #(w0 )+1 · . . . · pk #(wk )+1 , where p0 , p1 , . . . are the prime numbers 2,3,5,. . . in increasing order and #(=) = 0, #(∨) = 4, #(ci ) = 4 + 4i,

#(() = 1, #()) = 2, #(¬) = 3, #(∧) = 5, #(∃) = 6, #(∀) = 7, #(xi ) = 5 + 4i, #(Ri ) = 6 + 4i, #( f i ) = 7 + 4i.

Model theory

104

Table 6.1. Symbol

Interpretation in Mω

POS-IDx0 x1 x2

x0 is t = t  , where x1 = t and x2 = t   x0 is ¬t = t  , where x1 = t and x2 = t   x0 is Ri t1 . . . tri , where x1 = t1 , . . . , xri = tri  x0 is ¬Ri t1 . . . tri , where x1 = t1 , . . . , xri = tri  x0 is (φ ∧ ψ), where x1 = φ and x2 = ψ x0 is (φ ∨ ψ), where x1 = φ and x2 = ψ x0 is ∃xn φ, where x1 = n and x2 = φ x0 is ∀xn φ, where x1 = n and x2 = φ

NEG-IDx0 x1 x2 POS-ATOMi x0 x1 . . . xri NEG-ATOMi x0 x1 . . . xri CONJx0 x1 x2 DISJx0 x1 x2 EXIx0 x1 x2 UNIx0 x1 x2

The vocabulary L {+,×} has a symbol 0 for zero, a symbol 1 for one, and the names n of the other natural numbers n are defined inductively as terms +n1. We assume L {+,×} uses the symbols listed in Table 6.1 to represent syntactic operations: We assume that among the symbols of L {+,×} are functions that provide a bijection between elements of the model and finite sequences of elements of the model.2 Thus it makes sense to interpret arbitrary elements of Mω as assignments. We also assume that L {+,×} has the symbols given in Table 6.2 (we think of x0 as an assignment). All the above symbols are easily definable in terms of + and · in first order logic in any model Mω of Peano’s axioms, if wanted. Now we take a new predicate symbol SAT, not to be included in L {+,×} (and not to be definable in terms of + and · in first order logic) with the following intuitive meaning: SATx0 x1

2

x0 is an assignment s and x1 is φ for some L-formula φ such that Mω |=s φ.

In the special case that N Mω (i.e. N) is the whole universe of Mω , the encoding of finite sequences of elements of the model by elements of the model can be achieved by means of the unique factorization of integers, or alternatively by means of the Chinese Remainder Theorem.

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105

Table 6.2. Symbol

Interpretation in Mω

TRUE-IDx0 x1 x2

x0 satisfies the identity t = t  , where x1 = t and x2 = t   x0 satisfies the non-identity ¬t = t  , where x1 = t and x2 = t   x0 satisfies Ri t1 . . . tri , where x1 = t1 ,. . ., xri = tri  x0 satisfies ¬Ri t1 . . . tri , where x1 = t1 ,. . ., xri = tri  x0 and x2 are assignments that agree about variables other than x1

FALSE-IDx0 x1 x2 TRUE-ATOMi x0 x1 . . . xri FALSE-ATOMi x0 x1 . . . xri AGRx0 x1 x2

The point is that SAT is (implicitly) definable in terms of the others by the first order sentence θ L as follows: ∀x0 ∀x1 (SATx0 x1 ↔ ∃x2 ∃x3 (POS-IDx1 x2 x3 ∧ TRUE-IDx0 x2 x3 ) ∨∃x2 ∃x3 (NEG-IDx1 x2 x3 ∧ FALSE-IDx0 x2 x3 ) ∨∃x2 . . . ∃xr1 +1 (POS-ATOM1 x1 x2 . . . xr1 +1 ∧ TRUE-ATOM1 x0 x2 . . . xr1 +1 ) ∨... ∃x2 . . . ∃xrm +1 (POS-ATOMm x1 x2 . . . xrm +1 ∧ TRUE-ATOMm x0 x2 . . . xrm +1 ) ∨∃x2 . . . ∃xr1 +1 (NEG-ATOM1 x1 x2 . . . xr1 +1 ∧ FALSE-ATOM1 x0 x2 . . . xr1 +1 ) ∨... ∃x2 . . . ∃xrm +1 (NEG-ATOMm x1 x2 . . . xrm +1 ∧ FALSE-ATOMm x0 x2 . . . xrm +1 ) ∨∃x2 ∃x3 (CONJx1 x2 x3 ∧ (SATx0 x2 ∧ SATx0 x3 )) ∨∃x2 ∃x3 (DISJx1 x2 x3 ∧ (SATx0 x2 ∨ SATx0 x3 )) ∨∃x2 ∃x3 (EXIx1 x2 x3 ∧ ∃x4 (AGRx0 x2 x4 ∧ SATx4 x3 )) ∨∃x2 ∃x3 (UNIx1 x2 x3 ∧ ∀x4 (AGRx0 x2 x4 → SATx4 x3 ))).

The implicit (or inductive) nature of this definition (known as Tarski’s Truth Definition) manifests itself in the fact that SAT appears on both sides of the equivalence sign in θ L , and only positively in each case. There is no guarantee that θ really fixes what SAT is (see, e.g., ref. [26]). However, for the part of the actual formulas, or their representatives in Mω , the set SAT is unique. Theorem 6.21 If the L-structure Mω is a model of Peano’s axioms, then: (i) (Mω , SatN ) |= θ L , where SatN is the set of pairs s, φMω  such that Mω |=s φ; (ii) if (Mω , S) |= θ L and (Mω , S  ) |= θ L , then S ∩ SatN = S  ∩ SatN .

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Proof Claim (i) is tedious but trivial, assuming that our concepts are correctly defined. The claim we prove is the second one. To this end, suppose (Mω , S) |= θ L and (Mω , S  ) |= θ L . We prove s, φMω  ∈ S if and only if Mω |=s φ

(6.9)

for all φ. Since the same holds for S  by symmetry, we obtain the desired result. Case (1) φ is of the form t = t  , ¬t = t  , Ri t1 . . . tn or ¬Ri t1 . . . tn . The claim in Eq. (6.9) follows from our interpretation of POS-ID, NEG-ID, POS-ATOM, NEG-ATOM,TRUE-ID, FALSE-ID, TRUE-ATOMi , and FALSE-ATOMi in Mω . Case (2) φ is of the form (φ0 ∧ φ1 ). The claim in Eq. (6.9) follows from the conjunction part of the definition of θ L and our interpretation of CONJ in Mω . Case (3) φ is of the form (φ0 ∨ φ1 ). The claim in Eq. (6.9) follows from the disjunction part of the definition of θ L and our interpretation of DISJ in Mω . Case (4) φ is of the form ∃xn ψ. The claim in Eq. (6.9) follows from the part of the existential quantifier of the definition of θ L and our interpretation of AGR and EXI in Mω . Case (5) φ is of the form ∀xn ψ. The claim in Eq. (6.9) follows from the part of the universal quantifier of the definition of θ L and our interpretation of AGR and UNI in Mω .  By means of the satisfaction relation SAT we can define truth by means of the following formula: TRUE(x0 ) = ∃x1 SATx1 x0 . Corollary 6.22 The following are equivalent for all first order sentences φ in the vocabulary L and any model Mω of Peano’s axiom: (i) Mω |= φ; (ii) (Mω , S) |= (θ L ∧ TRUE (φ)) for some S ⊆ N2 ; (iii) (Mω , S) |= (θ L → TRUE (φ)) for all S ⊆ N2 . Let us call a model Mω of Peano’s axioms standard if the interpretation of the predicate N of the vocabulary of arithmetic in Mω is the set of natural numbers, the interpretation of + is the addition of natural numbers, and the interpretation of × is the multiplication of natural numbers. Corollary 6.23 Suppose Mω is any standard model of Peano’s axioms. If (Mω , S) |= θ L and (Mω , S  ) |= θ L , then S = S  .

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107

First order definable relations on standard Mω are called arithmetical. The definition of truth given by Corollary 6.22 is not first order, so we cannot say truth is arithmetical. A definition that has both existential second order and universal second order definition, as truth in the above corollary, is called hyperarithmetical. So we can say that first order truth on Mω is not arithmetical but hyperarithmetical. For more on hyperarithmetical definitions, see ref. [33]. Exercise 6.29 Prove Corollary 6.22. Exercise 6.30 Prove Corollary 6.23. Exercise 6.31 Show that we cannot remove the word “standard” from Corollary 6.23.

6.4.3 Definability of truth in D We now move from first order logic back to dependence logic. We observed in Corollary 6.22 that the truth definition of first order logic on a model Mω of Peano’s axioms can be given in first order logic if one existential second order quantifier is allowed. In dependence logic we can express the existential second order quantifier, and thus the truth definition of first order logic on Mω can be given in dependence logic. This can be extended to a truth definition of all of dependence logic, and this is our goal in this section. The fact that 11 has a truth definition in 11 on a structure with enough coding is well known in descriptive set theory. It was observed in ref. [19] that, as an application of Theorem 6.15, we have a truth definition for D in D on any structure with enough coding. We shall consider below a vocabulary L ⊇ L {+,×} ∪ {c}, where c is a new constant symbol. If φ is a sentence of D in this vocabulary, we indicate the inclusion of a new constant by writing φ as φ(c). Then, if d is another constant symbol, φ(d) is the sentence obtained from φ(c) by replacing c everywhere by d. Theorem 6.24 (ref. [19]) Suppose L ⊇ L {+,×} ∪ {c} is finite. There is a sentence τ (c) of D in the vocabulary L such that for all sentences φ of D in the vocabulary L and all models Mω of Peano’s axioms: Mω |= φ if and only if Mω |= τ (φ). Proof By Theorem 6.2, every sentence of D is logically equivalent to a 11 -sentence of following the form: ∃R1 . . . ∃Rn φ ∗ ,

(6.10)

Model theory

108

where φ ∗ is first order. We now replace the second order quantifiers ∃R1 . . . ∃Rn by just one second order quantifier ∃R0 , where R0 is a unary predicate symbol not in L. At the same time we replace every occurrence of Ri t1 . . . tn in φ ∗ by R0 i, t1 , . . . , tn , where (a1 , . . . , an ) → a1 , . . . , an  is the function in L {+,×} coding n-sequences. Let the result be φ ∗∗ . Now the following are equivalent for any φ in D in vocabulary L and any L-structure Mω , which is a model of Peano’s axioms: (i) Mω |= φ. (ii) (Mω , Z ) |= φ ∗∗ for some Z ⊆ N. Let L  = L ∪ {R0 }. By Corollary 6.22, we obtain the equivalence of (ii) with (iii) (Mω , Z , Sat) |= (θ L  ∧ TRUE(φ ∗∗ )) for some Sat ⊆ N2 and some Z ⊆ N. By Corollary 6.16, there is a sentence τ0 (c) of vocabulary L such that (iii) is equivalent to (iv) Mω |= τ0 (φ ∗∗ ). Let t(x0 ) be a term of the vocabulary L {+,×} such that for all sentences φ of D of vocabulary L the value of t(φ) in any L {+,×} -model of Peano’s axioms is φ ∗∗ . Let τ (c) be the L-sentence τ0 (t(c)). Then (i)–(iv) are equivalent to (v) Mω |= τ (φ).



Now that we have constructed the truth definition by recourse to 11 sentences, it should be pointed out that we could write τ (c) also directly in D, imitating the inductive definition of truth given in Definition 3.5. This is the approach taken in a forthcoming publication by Hodges and V¨aa¨ n¨anen.3 Let us now go back to the Liar Paradox. By Theorem 6.19, there is a sentence λ of D in the vocabulary L {+,×} such that for all Mω Mω |= λ if and only if Mω |= ¬τ (λ). Intuitively, λ says “This sentence is not true.” By Theorem 6.24, Mω |= λ if and only if Mω |= τ (λ). Thus Mω |= τ (λ) if and only if Mω |= ¬τ (λ), 3

W. Hodges and J. V¨aa¨ n¨anen, Dependence of variables construed as an atomic formula; to appear.

6.4 Truth definitions

109

which is, of course, only possible if Mω |= τ (λ) and Mω |= ¬τ (λ). Still another way of putting this is: Mω |= τ (λ) ∨ ¬τ (λ). Thus we get the pleasing result that the assertion that the Liar sentence is true (in the sense of τ (c)) is non-determined. This is in harmony with the intuition that the Liar sentence does not have a truth value. Exercise 6.32 Suppose Mω |= ψ0 if and only if Mω |= τ (ψ1 ) and Mω |= ψ1 if and only if Mω |= ¬τ (ψ0 ). Show that τ (ψ0 ) and τ (ψ1 ) are non-determined. Exercise 6.33 Contemplation of the sentence “This sentence is not true” leads immediately to the paradox ( ): the sentence is true if and only if it is not true. We have seen that we can write a sentence λ with the meaning “This sentence is not true” in D. Still we do not get the result that (λ ↔ ¬λ) is true. Indeed, show that (φ ↔ ¬φ) has no models, whatever φ in D is. Explain why the existence of λ does not lead to the paradox ( ). Exercise 6.34 Suppose ψ says “If ψ is true, then ¬ψ is true,” i.e. Mω |= ψ if and only if Mω |= (τ (ψ) → τ (¬ψ)). Show that ψ is non-determined. Exercise 6.35 Suppose ψ says “It is not true that ψ is true,” i.e. Mω |= ψ if and only if Mω |= ¬τ (τ (ψ)). Show that τ (τ (ψ)) is non-determined. Exercise 6.36 Show that there cannot be τ  (c) in D such that for all φ and all Mω we have Mω |= φ if and only if Mω |= τ  (φ). Exercise 6.37 (See ref. [35].) Suppose Mω is a model of Peano’s axioms, as above. Let T be the set of sentences of D that are true in Mω . Show that there

110

Model theory

is a model M of T such that for some a ∈ M we have M |= (τ (a) ∧ τ (¬a)).

6.5 Model existence game In this section we learn a new game associated with trying to construct a model for a sentence or a set of sentences. The new game, called the model existence game, is like H (φ) except that there is no model present. So the game is about an imagined model. In particular, player II claims she has a model for φ but she does not tell player I what it is. It turns out that if she has a winning strategy, then there actually is a model for φ and we can build it from the winning strategy.

6.5.1 First order case We first treat the case of first order logic. After this we shall easily extend the method to all of dependence logic. In the model existence game we talk about semantics without actually having any model. Let us take a countably infinite set C of new constant symbols. Intuitively the new constants c are elements of the imaginary model. Definition 6.25 The model existence game MEG (T, L) of the set T of first order L-sentences is defined as follows. Let C be a countably infinite set of new constant symbols. Let L  = L ∪ C. There are two players, I and II. We denote players by α, and the opponent of α by α ∗ . A position of the game is a pair (φ, α), where φ is an L  -sentence and α ∈ {I, II}. At the beginning of the game the position is (, II). The rules of the game are as follows. Player I chooses φ ∈ T and the game continues from the position (φ, II). (2) Identity move Player I chooses a constant L  -term t. The game continues from the position (t = t, II). (3) Substitution move Player I chooses previous positions (φ(t), α) and (t = t  , II), where φ(t) is atomic, and then the game continues from the position (φ(t  ), α). (4) Negation move Player I chooses a previous position (¬φ, α). Then the game continues from the position (φ, α ∗ ). (5) Disjunction move Player I chooses a previous position (φ0 ∨ φ1 , α). Then player α decides whether the game continues from the position (φ0 , α) or from the position (φ1 , α).

(1) Theory move

6.5 Model existence game

111

Table 6.3. I

Pc ∨ ¬P f c ¬P f c

II

Rule

∃x0 ¬(P x0 ∨ ¬P f x0 ) ¬(Pc ∨ ¬P f c)

(1) (6) (4) (5) (4) (5) (7) (3) (7) (7)

Pfc Pc

(6) Existential move

(7) Constant move

f c = c Pc f c = c f c = c ... II wins

Player I chooses a previous position of the form (∃xn φ(xn ), α). Player α chooses c ∈ C and the game continues from the position (φ(c), α). Player I chooses a constant L  -term t. Then player II chooses c ∈ C and the game continues from the position (t = c, II).

Player I wins if during the game both (φ, II) and (φ, I) occur as positions for some atomic L  -sentence φ. Otherwise II wins. Example 6.26 Player II has a winning strategy in MEG({∃x0 ¬(P x0 ∨ ¬P f x0 )}, {P, f }). Table 6.3 shows a play of the game, when II plays her winning strategy. The rules would allow I to prolong the game by repeating previous moves but II would always give the same responses, and in the end II wins anyway. Example 6.27 In this example, I has a winning strategy in MEG({¬∃x0 (Pd ∨ ¬P x0 )}, {P, d}) (see Table 6.4). A winning strategy of II in MEG(T, L) can be conveniently presented in the form of a set of sets. Definition 6.28 Let L be a countable vocabulary, let C be a countable set of new constant symbols, and let L  = L ∪ C. A consistency property is any set  of finite sets S of pairs (φ, α), where φ is a first order L-formula, and α ∈ {I, II}, which satisfies the following conditions.

Model theory

112

Table 6.4. I ∃x0 (Pd ∨ ¬P x0 ) Pd ∨ ¬Pc Pd Pc ¬Pc

II

Rule

¬∃x0 (Pd ∨ ¬P x0 )

(1) (4) (7) (6) (5) (3) (5) (4)

d=c

Pc I has won

(i) If S ∈ , then S ∪ {(t = t, II)} ∈  for every constant L  -term t. (ii) If (φ(t), α) ∈ S ∈ , φ(t) atomic, and (t = t  , II) ∈ S, then S ∪ {(φ(t  ), α)} ∈ . (iii) If (¬φ, α) ∈ S ∈ , then S ∪ {(φ, α ∗ )} ∈ . (iv) If (φ ∨ ψ, II) ∈ S ∈ , then S ∪ {(φ, II)} ∈  or S ∪ {(ψ, II)} ∈ . (v) If (φ ∨ ψ, I) ∈ S ∈ , then S ∪ {(φ, I)} ∈  and S ∪ {(ψ, I)} ∈ . (vi) If (∃xn φ(xn ), II) ∈ S ∈ , then S ∪ {(φ(c), II)} ∈  for some c ∈ C. (vii) If (∃xn φ(xn ), I) ∈ S ∈ , then S ∪ {(φ(c), I)} ∈  for all c ∈ C. (viii) For every constant L  -term t there is c ∈ C such that S ∪ {(t = c, II)} ∈ . (ix) There is no atomic formula φ such that (φ, II) ∈ S and (φ, I) ∈ S. The consistency property  is a consistency property for a set T of first order L-sentences if for all S ∈  and all φ ∈ T we have S ∪ {(φ, II)} ∈ . Lemma 6.29 The following are equivalent. (i) Player II has a winning strategy in MEG(T, L). (ii) There is a consistency property  for T . Proof Suppose σ is a winning strategy of II in MEG(T, L). Suppose the game MEG(T, L) is played for n rounds and II plays σ . A certain set S of positions (φ, α) is generated. Let  be the set of all possible such sets S, when n ∈ N. Clearly  is a consistency property and for all S ∈  and all φ ∈ T we have S ∪ {(φ, II)} ∈ . Conversely, if such a consistency property  existed, player II would win MEG(T, L) by making sure the set of positions played is in . 

6.5 Model existence game

113

An extreme case of a consistency property  is the situation in which all the finite sets in  are actually subsets of one and the same set H , and this set H has nice closure properties (see below). Such sets are called Hintikka sets and they were introduced in ref. [18]. In many applications of the model existence game such a set is all we need. More refined proofs use a consistency property. Definition 6.30 Let L be a let countable vocabulary, let C be a countable set of new constant symbols, and let L  = L ∪ C. A Hintikka set is any set H of pairs (φ, α), where φ is a first order L  -sentence and α ∈ {I, II}, which satisfies the following conditions. (t = t, II) ∈ H for every constant L  -term t. If (φ(t), α) ∈ H , φ(t) atomic, and (t = t  , II) ∈ H , then (φ(t  ), α) ∈ H . If (¬φ, α) ∈ H , then (φ, α ∗ ) ∈ H . If (φ ∨ ψ, II) ∈ H , then (φ, II) ∈ H or (ψ, II) ∈ H . If (φ ∨ ψ, I) ∈ H , then (φ, I) ∈ H and (ψ, I) ∈ H . If (∃xn φ(xn ), II) ∈ H , then (φ(c), II) ∈ H for some c ∈ C If (∃xn φ, I) ∈ H , then (φ(c), I) ∈ H for all c ∈ C For every constant L  -term t and every (φ, α) ∈ H there is c ∈ C such that (t = c, II) ∈ H . (ix) There is no atomic sentence φ such that (φ, II) ∈ H and (φ, I) ∈ H .

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

The Hintikka set H is a Hintikka set for a set T of first order L-sentences if (φ, II) ∈ H for all φ ∈ T . Lemma 6.31 establishes the basic connection between consistency properties and Hintikka sets. Roughly speaking, Hintikka sets are unions of increasing chains in a consistency property, but this is not quite true: the increasing chains have to be constructed carefully. Lemma 6.31 Let T be a set of first order L-sentences. (i) Suppose  is a consistency property for T . Then there is an increasing  sequence of Sn ∈  such that H = n Sn is a Hintikka set for T . (ii) If H is a Hintikka set for T , then  = {S ⊆ H : S is finite} is a consistency property for T . Proof To prove the first claim, let T r m be the set of all constant L  -terms. Let T = {φn : n ∈ N}; C = {cn : n ∈ N}; T r m = {tn : n ∈ N}.

114

Model theory

Let { pn : n ∈ N} be a list of all elements of . We let S0 ∈  be arbitrary and define an increasing sequence Sn ∈  as follows: (i) If n = 3i , then Sn+1 is Sn ∪ {(φi , II)} ∈ . (ii) If n = 2 · 3i , then Sn+1 is Sn ∪ {(ti = ti , II)} ∈ . (iii) If n = 4 · 3i · 5 j , pi = (t = t  , II) ∈ Sn , and p j = (φ(t), α) ∈ Sn with φ(t) atomic, then Sn+1 is Sn ∪ {(φ(t  ), α)} ∈ . (iv) If n = 8 · 3i and pi = (¬ψ, α) ∈ Sn , then Sn+1 is Sn ∪ {(ψ, α ∗ )}. (v) If n = 16 · 3i and pi = (θ0 ∨ θ1 , II) ∈ Sn , then Sn+1 is Sn ∪ {(θ1 , II)} or Sn ∪ {(θ1 , II)}, whichever is in . (vi) If n = 32 · 3i · 5 j , j ∈ {0, 1} and pi = (θ0 ∨ θ1 , I) ∈ Sn , then Sn+1 is Sn ∪ {(θ j , I)} ∈ . (vii) If n = 64 · 3i and pi = (∃xk φ, II) ∈ Sn , then Sn+1 is Sn ∪ {(φ(c), II)} for such c ∈ C that Sn+1 ∈ . (viii) If n = 128 · 3i · 5 j and pi = (∃xk φ, I) ∈ Sn , then Sn+1 is Sn ∪ {(φ(c j ), I)} ∈ . (ix) If n = 256 · 3i , then Sn+1 is Sn ∪ {(ti = c, II)} for such c ∈ C that Sn+1 ∈ .  Clearly n Sn is a Hintikka set for T . The second claim is trivial.  Combining Lemmas 6.29 and 6.31 immediately yields the following. Corollary 6.32 The following are equivalent. (i) There is a Hintikka set H for T . (ii) Player II has a winning strategy in MEG(T, L) Theorem 6.33 (Model Existence Theorem for first order logic) Suppose L is a countable vocabulary and T is a set of L-sentences. The following are equivalent. (i) There is an L-structure M such that M |= T . (ii) Player II has a winning strategy in MEG(T, L). Proof Let us first assume such an M exists. The winning strategy of player II in MEG(T, L) is to maintain the condition ( ) that if the position is (φ, α), then (φ, ∅, α) ∈ T . Player II defines interpretations of the constants c ∈ C in M during the game, step by step, as soon as they appear. (1) Position (φ, ∅, II), where φ ∈ T . Then (φ, ∅, II) ∈ T , since M is a model of T .

6.5 Model existence game

115

(2) Position (t = t, II). Of course, M |= t = t, if the constants of t are interpreted in M. If they are not, we can interpret them in an arbitrary way. (3) Player I has chosen previous positions (φ(t), α) and (t = t  , II), and then the game has continued to the position (φ(t  ), α). By the induction hypothesis, M |= t = t  . Thus (φ(t), ∅, α) ∈ T implies (φ(t  ), ∅, α) ∈ T . (4) Position (¬φ, α). Since (¬φ, ∅, α) ∈ T , we have (φ, ∅, α ∗ ) ∈ T . Thus II can play this move according to her plan. (5) Position (φ ∨ ψ, α). We know (φ ∨ ψ, ∅, α) ∈ T . Let us first assume α = II. By the definition of T , (φ, ∅, II) or (ψ, ∅, II) is in T . Thus player II can choose one of (φ, II), (ψ, II), and the respective triple (φ, ∅, II) or (ψ, ∅, II) is in T . Condition ( ) remains valid. Let us then assume α = I. Whichever choice player I makes in position (φ ∨ ψ, I), condition ( ) remains valid, as both (φ, ∅, I) and (ψ, ∅, I) are in T . (6) Position (∃xn φ(xn ), α). We know (∃xn φ(xn ), ∅, α) ∈ T . Let us first assume α = II. By the definition of T , there is a ∈ M such that (φ, {(n, a)}, II) is in T . Let us choose a new constant symbol c and interpret it in M as a. Now the game can proceed to (φ(c), II) and condition ( ) remains valid. Suppose then α = I and I chooses c ∈ C. Now (φ, {(n, cM )}, α) ∈ T , whatever cM ∈ M is. If cM is defined already, we are done, otherwise we let cM be an arbitrary4 element of M. Thus condition ( ) remains valid. (7) Player I chooses a constant L  -term t. If some constants of t are not interpreted yet in M, we interpret them in an arbitrary way (it does not matter which value they are given). Then a = t M is defined. Now player II chooses any new c ∈ C and the game continues to the position (t = c, II). We let cM = a, and condition ( ) remains valid. Let us then observe that if II plays according to this plan, she wins, for if she ends up playing both (ψ, II) and (ψ, I) for some atomic L  -sentence ψ, then {(ψ, ∅, II) ∈ T and (ψ, ∅, I)} ∈ T , a contradiction. On the other hand, suppose player II has a winning strategy in the game MEG(T, L). By Corollary 6.32, there is a Hintikka set H such that (φ, II) ∈ H for all φ ∈ T . Define c ∼ c if (c = c , II) ∈ H . The relation ∼ is an equivalence relation on C. Let us define an L-structure M as follows. We let M = {[c] : c ∈ C}. If d ∈ L is a constant symbol, we choose c ∈ C such that (d = c, II) ∈ H and define d M = [c]. If f ∈ L and #( f ) = n we let f M ([ci1 ], . . . , [cin ]) = [c] for some c such that ( f ci1 . . . cin = c, II) ∈ H . For any term t and any u there is a c ∈ C such that (t = c, II) ∈ H . It is easy to see that t M = [c]. For a relation symbol R we let [c1 ], . . . , [cn ] ∈ R M if and only if (Rc1 . . . cn , II) is in H . 4

To be precise, we can choose once and for all an element a0 of M always to be used at this step.

Model theory

116

We prove by induction on φ that if φ is an L-formula then (

)

If (φ, α) ∈ H, then(φ, ∅, α) ∈ T .

(i) φ is an equation t = t  . If (t = t  , II) ∈ H , then there are c and c in C such that (t = c, II), (t  = c) ∈ H , and hence t M = [c] = [c ] = (t  )M . Thus (t = t  , ∅, II) ∈ T . Similarly, if (t = t  , I) ∈ H , then (t = t  , II) ∈ / H , whence [c] = [c ], and (t = t  , ∅, I) ∈ T follows. (ii) φ is an atomic sentence Rt1 . . . tn . If (Rt1 . . . tn , II) ∈ H , then there are c1 , . . . , cn in C such that (ti = ci , II) ∈ H for i = 1, . . . , n, whence (Rc1 . . . cn , II) ∈ H , and hence by the definition of M, ([c1 ], . . . , [cn ]) ∈ R M . Thus (Rt1 . . . tn , ∅, II) ∈ T . If (Rt1 . . . tn , I) ∈ H , then the pair (Rc1 . . . cn , II) is not in H , and hence by the definition of M, ([c1 ], . . . , [cn ]) ∈ / R M . Thus (Rt1 . . . tn , ∅, I) ∈ T . (iii) Negation: if (¬φ, α) ∈ H , then (φ, α ∗ ) ∈ H , hence by the induction hypothesis (φ, ∅, α ∗ ) ∈ T , i.e. (φ, ∅, α) ∈ T . (iv) Disjunction: if (φ ∨ ψ, II) ∈ H , then (φ, II) ∈ H or (ψ, II) ∈ H , hence by the induction hypothesis (φ, ∅, II) ∈ T or (ψ, ∅, II) ∈ T , i.e. (φ ∨ ψ, ∅, II) ∈ T . If (φ ∨ ψ, I) ∈ H , then (φ, I) ∈ H and (ψ, I) ∈ H , hence by the induction hypothesis (φ, ∅, I) ∈ T and (ψ, ∅, I) ∈ T , i.e. (φ ∨ ψ, ∅, I) ∈ T . (v) Quantifier: if (∃xn φ(xn ), II) ∈ H , there is a c ∈ C such that the pair (φ(c), II) is in H . By the induction hypothesis (φ(c), ∅, II) ∈ T . Thus the triple (φ, {(n, [c])}, II) is in T , whence the triple (∃xn φ(xn ), ∅, II) is in T . If the pair (∃xn φ(xn ), I) is in H , and [c] ∈ M is arbitrary, then (φ(c), I) ∈ H . By the induction hypothesis (φ(c), ∅, I) ∈ T . Thus (φ, {(n, [c])}, I) ∈ T , whence (∃xn φ(xn ), ∅, I) ∈ T . In particular, since H is a Hintikka set for T , M |= T .



Corollary 6.34 Suppose L is a countable vocabulary, T is a set of L-sentences, and φ is any L-sentence. Then the following are equivalent: (i) T |= φ; (ii) player I has a winning strategy in MEG(T ∪ {¬φ}, L). Proof The game MEG(T ∪ {¬φ}, L) is determined. So, by Theorem 6.33, condition (ii) is equivalent to T ∪ {¬φ} not having a model, which is exactly what condition 1 says (remember that φ is assumed to be first order).  In first order logic, condition (i) of Corollary 6.34 is equivalent to φ having a formal proof from T (see, e.g., ref. [8] for a definition of formal proof). We can think of a winning strategy of player I in MEG(T ∪ {¬φ}, L) as a semantic

6.5 Model existence game

117

proof of φ from T . In the literature this concept, or its close relatives, occur under the name of semantic tree or Beth tableaux. The model existence theorem has many applications in first order logic. We mention, in particular, the Compactness Theorem, the Omitting Types Theorem and the Interpolation Theorem.

6.5.2 Model existence game for Skolem Normal Form Suppose φ is a 11 -sentence in Skolem normal form, ∃ f 1 . . . ∃ f n ∀x1 . . . ∀xm ψ,

(6.11)

where ψ is quantifier-free and f 1 , . . . , f n are function symbols not in L. We denote the extension of L by the various new function symbols f 1 , . . . , f n occurring in the formulas in Eq. (6.11) by L ∗ . The model existence game that we now define for 11 is shockingly simple in that its positions involve quantifier-free sentences only. Definition 6.35 The model existence game MEG11 (T, L) of the set T of 11 sentences in Skolem normal form of the vocabulary L is defined as follows. Let C be a countably infinite set of new constant symbols. Let L  = L ∗ ∪ C. A position of the game is a pair (φ, α), where φ is a quantifier-free first order L  -formula and α ∈ {I, II}. At the beginning of the game, the position is (, II). The rules of the game are as follows. (1) Theory move.

Player I chooses ∃ f 1 . . . ∃ f n ∀xi1 . . . ∀xim ψ(xi1 , . . . , xim ) ∈ T

and c1 , . . . , cn ∈ C, and then the game continues from the position (ψ(c1 , . . . , cn ), II). (2) Identity move. Player I chooses an L  -term t. The game continues from the position (t = t, II). (3) Substitution move. Player I chooses previous positions (φ(t), α), φ(t) atomic, and (t = t  , II), and then the game continues from the position (φ(t  ), α). (4) Negation move. Player I chooses a previous position (¬φ, α). Then the game continues from the position (φ, α ∗ ). (5) Disjunction move. Player I chooses a previous position (φ0 ∨ φ1 , α). Then player α decides whether the game continues from the position (φ0 , α) or from the position (φ1 , α).

118

(6) Constant move.

Model theory Player I chooses an L  -term t. Then player II chooses c ∈ C and the game continues from the position (t = c, II).

Player I wins if at any point II has played (φ, II) and (φ, I) for some atomic L  -sentence φ. Otherwise II wins. Note that I can play the Theory move several times during the game; indeed, he typically may play it infinitely many times. The following result follows immediately from Theorem 6.33. Theorem 6.36 (Model Existence Theorem for Skolem Normal Form) Suppose L is a countable vocabulary and T is a set of 11 -sentences of vocabulary L in Skolem Normal Form. The following are equivalent. (i) There is an L-structure M such that M |= T . (ii) Player II has a winning strategy in MEG11 (T, L). Since 11 is not closed under negation, we do not get a concept of “semantic proof” for 11 from MEG11 (T, L ∗ ). It does not make sense to talk about MEG11 (T ∪ {¬φ}, L ∗ ), since we cannot in general write ¬φ in Skolem Normal Form, unless it is actually first order. Instead we can view Theorem 6.36 as a useful criterion of consistency of a set of 11 sentences. So the model existence game provides 11 a criterion for consistency but no criterion of logical consequence and no concept of proof. We know from Theorem 6.15 that every sentence φ of dependence logic is logically equivalent to a 11 -sentence τ1,φ . Moreover, also ¬φ is logically equivalent to a 11 -sentence τ0,φ . If T is a set of sentences of dependence logic, let τT be the set of all τ1,φ , where φ ∈ T . Thus we obtain from the above the following. Corollary 6.37 Suppose L is a countable vocabulary, T is a set of sentences of dependence logic of vocabulary L, and φ is any sentence of dependence logic in vocabulary L. Then the following are equivalent. (i) No model of T is a model of ¬φ. (ii) Player I has a winning strategy in MEG11 (τT ∪ {τ0,φ }, L). Unfortunately, condition (i) in Corollary 6.37 is not equivalent to T |= φ, as φ may be non-determined in some models of T . Certainly, if T |= φ, condition (i) holds. So, condition (i) is a weak form of provability for D. It would be strong enough to give T |= φ if φ was determined, but then φ would be first order. As it is, condition (ii) gives a complete criterion for deciding whether a first order sentence is a logical consequence of a set of sentences of D.

6.5 Model existence game

119

Exercise 6.38 Describe a winning strategy of player II in MEG(T, L), when. (i) T = {∀x0 (P x0 ∨ Qx0 ), ∃x0 P x0 , ∃x0 ¬Qx0 }; (ii) T = {∀x Rx f x, ¬∃x∀y¬Rx y}. Exercise 6.39 Describe a winning strategy of player I in the game MEG(T, L), when (i) T = {∀x0 (P x → Qx), ∃x0 P x0 , ¬∃x0 Qx0 }; (ii) T = {∀x Rx f x, ¬∀x∃y Rx y}. Exercise 6.40 Let L be let a countable vocabulary, let C be a countable set of new constant symbols and L  = L ∪ C. Let  be the set of finite sets S of L  -sentences such that for each S there is a model M S in which every element is an interpretation of some constant c ∈ C, and which satisfies (φ, II) ∈ S implies M S |= φ, and (φ, I) ∈ S implies M S |= ¬φ. Show that  is a consistency property. Exercise 6.41 Let Mω be a model of Peano’s axioms. Let T be the set of first order L {+,×} -sentences that are true in Mω . Let d be a new constant symbol and L = L {+,×} ∪ {d}. Let T  be the union of T and the sentences n < d for all n < ω. Show that II can win the game MEG(T  , L) with the following strategy: if the played positions are (φi , αi ), i = 1, . . . , n, then Mω has an expansion that is a model of {φi : αi = II} ∪ {¬φi : αi = I}. Exercise 6.42 Prove in detail that (i) → (ii) in Lemma 6.29. Exercise 6.43 Prove in detail that (ii) → (i) in Lemma 6.29. Exercise 6.44 Suppose H is a Hintikka set. Define c ∼ c if (c = c , II) ∈ H . Prove that ∼ is an equivalence relation on C. Exercise 6.45 Suppose S1 and S2 are sets of first order sentences. Let the vocabulary of S1 be L 1 and let the vocabulary of S2 be L 2 . If L ⊆ L i , we denote by Si L the set of φ ∈ Si , the vocabulary of which is contained in L. We say that an L 1 ∩ L 2 -sentence θ separates S1 and S2 if S1 |= θ and S2 ∪ {θ } has no models, and that S1 and S2 are separable if some first order θ separates them; otherwise they are inseparable. Suppose L 1 and L 2 are vocabularies consisting (for simplicity) of relation symbols only. Suppose φ is a first order L 1 -sentence and ψ is a first order L 2 -sentence. Let C be a countably infinite set of new constant symbols. Let  be the set of all finite sets S of pairs (θ, α) satisfying the following conditions.

120

Model theory

(i) If (θ, α) ∈ S, then θ is an L 1 ∪ C-sentence or an L 2 ∪ C-sentence and α ∈ {I, II}. (ii) If T (S) = {θ : (θ, II) ∈ S} ∪ {¬θ : (θ, I) ∈ S}, then T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C) are inseparable. Show that  is a consistency property if conditions (i) and (viii) of Definition 6.28 are modified to refer only to terms t that are either L 1 ∪ C-terms or L 2 ∪ C-terms. (This exercise can be used to prove the Separation Theorem (Theorem 6.7).) Exercise 6.46 Let T be a set of sentences of D of a countable vocabulary L such that every finite subset of T has a model. Describe, without appealing to the Compactness Theorem (Theorem 6.4), a winning strategy of II in MEG11 (τT , L). This, combined with the Model Existence Theorem (Theorem 6.36) gives a proof of the Compactness Theorem (Theorem 6.4). Exercise 6.47 Give sentences φ and ψ of D such that φ and ψ have no finite models in common, and, if we denote the vocabulary of φ by L and the vocabulary of ψ by L  , then there is no first order sentence θ of vocabulary L ∩ L  such that every finite model of φ is a model of θ , and θ and ψ have no finite models in common. Exercise 6.48 Suppose we have a countable vocabulary L and T is a set of first order L-sentences. A type of T is a countable sequence ψ0 (x0 ), ψ1 (x0 ), . . . of formulas with just one free variable x such that for all n: T ∪ {∃x(ψ0 (x) ∧ . . . ∧ ψn (x))} has a model. A model M of T satisfies a type p if some element a of M satisfies M |={(x0 ,a)} ψ(x0 ) for all ψ(x0 ) ∈ p. A model M of T omits the type p if no element of M satisfies p. A type p of T is principal if there is a formula θ (x0 ) such that r T |= θ (x ) → ψ(x ) for all ψ(x ) ∈ p; 0 0 0 r every finite subset of T ∪ {∃x θ (x )} has a model. 0 0 Suppose p = {ψn (x0 ) : n ∈ N} is a non-principal type of T . By Exercise 6.46, player II has a winning strategy in the game MEG(T, L). Let  be a consistency property for T . Show that there is a function f : N → N and an increasing  sequence Sn in  such that n Sn is a Hintikka set for T ∪ {¬φ f (n) (cn ) : n ∈ N}. Derive the Omitting Types Theorem: if L is a countable vocabulary, T a set of L-sentences such that T has a model, and p a non-principal type of T , then T has a countable model which omits p. Exercise 6.49 (Failure of Omitting Types Theorem in D) This is a continuation of Exercise 6.48. Give an example of a first order theory T in a countable

6.6 Ehrenfeucht–Fra¨ıss´e game for dependence logic

121

vocabulary L and type p = {ψn (x0 ) : n ∈ N} of formulas of D in the vocabulary L such that r p is a non-principal type of T in the sense of Exercise 6.48; r no model of T omits p. Exercise 6.50 Let us write T D φ if player I has a winning strategy in MEG11 (τT ∪ {τ0,φ }, L), or equivalently, if no model of T is a model of ¬φ. Show that (i) if T D φ and T D ψ, then T D (φ ∧ ψ); (ii) if φ D θ and ψ D θ, then (φ ∨ ψ) D θ; (iii) φ D ¬φ and ¬φ D φ. Exercise 6.51 Continuation of Exercise 6.50. Show that (i) T has a model if and only if T D ¬; (ii) T D (φ ∧ ¬φ) if and only if φ is non-determined in models of T . Exercise 6.52 Continuation of Exercise 6.50. Show that φ D ψ and ψ D θ do not necessarily imply φ D θ. Exercise 6.53 Continuation of Exercise 6.50. Give T and φ such that T D φ but there is no finite T0 ⊆ T such that T0 D φ.

6.6 Ehrenfeucht–Fra¨ıss´e game for dependence logic We define the concept of elementary equivalence of models and give this concept a game characterization. In this section we assume that vocabularies do not contain function symbols, for simplicity. Definition 6.38 Two models M and N of the same vocabulary are Dequivalent in symbols M ≡D N if they satisfy the same sentences of dependence logic. It is easy to see that isomorphic structures are D-equivalent and that Dequivalence is an equivalence relation. However, despite the quite strong L¨owenheim–Skolem Theorem of dependence logic, we have the following negative result about D-equivalence. Proposition 6.39 There is an uncountable model M in a finite vocabulary such that M is not D-equivalent to any countable models.

Model theory

122

M

N s

s

Fig. 6.1. A position in the ordinary Ehrenfeucht–Fra¨ıss´e game.

Proof Let M = (R, N) be a model for the vocabulary {P}. Suppose M ≡D N , where N is countable. It is clear that there is a one-to-one function from N into P N . Let φ be a sentence of D which says exactly this. Since N |= φ, we have M |= φ, a contradiction.  It turns out that the following, more basic, concept is a far better concept to start with. Definition 6.40 Suppose M and N are structures for the same vocabulary. We say that M is D-semiequivalent to N , in symbols MD N , if N satisfies every sentence of dependence logic that is true in M. Note that equivalence and semiequivalence are equivalent concepts if dependence logic is replaced by first order logic. Proposition 6.41 Every infinite model in a countable vocabulary is D-semiequivalent to models of all infinite cardinalities. Proof Suppose M is an infinite model with a countable vocabulary L. Let T be the set of sentences of dependence logic in the vocabulary L that are true in M. By Theorem 6.5 the theory T has models of all infinite cardinalities (Theorem 6.5 is formulated for sentences only, but the proof for countable theories is the same).  We now introduce an Ehrenfeucht–Fra¨ıss´e game adequate for dependence logic and use this game to characterize D . See Figs. 6.1 and 6.2. Definition 6.42 Let M and N be two structures of the same vocabulary. The game EFn has two players and n moves. The position after move m is a pair (X, Y ), where X ⊆ M m and Y ⊆ N m for some m. In the beginning, the position

6.6 Ehrenfeucht–Fra¨ıss´e game for dependence logic

M

123

N X Y

Fig. 6.2. A position in the Ehrenfeucht–Fra¨ıss´e game for D.

Y0

X0 X1 M

Y1 N

Fig. 6.3. A splitting move.

is ({∅}, {∅}) and i 0 = 0. Suppose the position after move number m is (X, Y ). There are the following possibilities for the continuation of the game. Splitting move. Player I represents X as a union X = X 0 ∪ X 1 . Then player II represents Y as a union Y = Y0 ∪ Y1 . Now player I chooses whether the game continues from the position (X 0 , Y0 ) or from the position (X 1 , Y1 ); see Fig. 6.3. Duplication move. Player I decides that the game should continue from the new position (X (M/xim ), Y (N /xim )); see Fig. 6.4. Supplementing move. Player I chooses a function F : X → M. Then player II chooses a function G : Y → N . Then the game continues from the position (X (F/xim ), Y (G/xim )); see Fig. 6.5.

Model theory

124

M

N X(M/ xn ) Y (N/ xn)

Fig. 6.4. A duplication move.

M

N X(F/xn)

Y (G/xn)

Fig. 6.5. A supplementing move.

After n moves the position (X n , Yn ) is reached and the game ends. Player II is the winner if M |= X n φ ⇒ N |=Yn φ holds for all atomic and negated atomic and dependence formulas of the form φ(x0 , . . . , xin −1 ). Otherwise player I wins. This is a game of perfect information, and the concept of winning strategy is defined as usual. By the Gale–Stewart Theorem, the game is determined. Definition 6.43 (i) (ii) (iii) (iv)

qr(φ) = 0 if φ is atomic or a dependence formula. qr(φ ∨ ψ) = max(qr(φ), qr(ψ)) + 1. qr(∃xn φ) = qr(φ) + 1. qr(¬φ) : (a) qr(¬φ) = 0 if φ is atomic or a dependence formula; (b) qr(¬¬φ) = qr(φ); (c) qr(¬(φ ∨ ψ)) = max(qr(¬φ), qr(¬ψ)); (d) qr(¬∃xn φ) = qr(¬φ) + 1.

6.6 Ehrenfeucht–Fra¨ıss´e game for dependence logic

125

Let Fmlm n be the set of formulas φ of D with qr φ ≤ m and with free variables among x0 , . . . , xn−1 . We write M nD N if M |= φ implies N |= φ for all φ in Fmln0 , and M ≡nD N if M |= φ is equivalent to N |= φ for all φ in Fmln0 . Note that there are for each n and m, up to logical equivalence, only finitely many formulas in Fmlm n. Theorem 6.44 Suppose M and N are models of the same vocabulary. Then the following are equivalent: (1) player II has a winning strategy in the game EFn (M, N ); (2) M nD N . Proof We prove the equivalence, for all n, of the following two statements. (3)m Player II has a winning strategy in the game EFm (M, N ) in position (X, Y ), where X ⊆ M n and Y ⊆ N n . (4)m If φ is a formula in Fmlm n , then M |= X φ ⇒ N |=Y φ.

(6.12)

The proof is by induction on m. For each m we prove the claim simultaneously for all n. The case m = 0 is true by construction. Let us then assume (3)m ⇐⇒ (4)m as an induction hypothesis. Assume now (3)m+1 and let φ be a formula in Fmlm+1 such that M |= X φ. As part of the induction hypothesis, n we assume that the claim in Eq. (6.12) holds for formulas shorter than φ. Case (1) φ = ψ0 ∨ ψ1 , where ψ0 , ψ1 ∈ Fmlm n . Since M |= X φ, there are X 0 and X 1 such that X = X 0 ∪ X 1 , M |= X 0 ψ0 and M |= X 1 ψ1 . We let I play {X 0 , X 1 }. Then II plays according to her winning strategy {Y0 , Y1 }. Since the next position in the game can be either one of (X 0 , Y0 ), (X 1 , Y1 ), we can apply the induction hypothesis to both. This yields N |=Y0 ψ0 and N |=Y1 ψ1 . Thus N |=Y φ. Case (2) φ = ∃xn ψ, where ψ ∈ Fmlm n−1 . Since M |= X φ, there is a function F : X → M such that M |= X (F/xn ) ψ. We let I play F. Then II plays according to her winning strategy a function G : Y → N and the game continues in position (X (F/xn ), Y (G/xn )). The induction hypothesis gives N |=Y (G/xn ) ψ. Now N |=Y φ follows. Case (3) φ = ¬¬ψ, n = max(n 0 , n 1 ). Since M |= X φ, we have M |= X ψ. By the induction hypothesis, N |=Y ψ. Thus N |=Y φ. Case (4) φ = ¬(ψ0 ∨ ψ1 ), where ψ0 , ψ1 ∈ Fmlm n . Since M |= X φ, we have M |= X ¬ψ0 and M |= X ¬ψ1 . By the induction hypothesis, N |=Y ¬ψ0 and N |=Y ¬ψ1 . Thus N |=Y φ.

Model theory

126

Case (5) φ = ¬∃xn ψ, where ¬ψ ∈ Fmlm n+1 . By assumption, M |= X (M/xn ) ¬ψ. We let now I demand that the game continues in the duplicated position (X (M/xn ), Y (N /xn )). The induction hypothesis gives N |=Y (N /xn ) ¬ψ. Now N |=Y φ follows trivially. To prove the converse implication, assume (4)m+1 . To prove (3)m+1 we consider the possible moves that player I can make in the position (X, Y ). Case (i) Player I writes X = X 0 ∪ X 1 . Let φ j , j < k, be a complete list (up to logical equivalence) of formulas in Fmlm n . Since  ¬φ j M |= X 0 ¬ M|= X 0 φ j

and



M |= X 1 ¬

¬φ j ,

M|= X 1 φ j

we have

⎛

⎞



M |= X ⎝¬

⎛

¬φ j ⎠ ∨ ⎝¬

M|= X 0 φ j

Note that   qr ¬ M|= X 1 φ j



⎞ ¬φ j ⎠ .

M|= X 1 φ j

 ¬φ j

= max (¬¬φ j ) = max φ j ≤ m. M|= X 1 φ j

M|= X 1 φ j

Therefore, by (4)m+1 , ⎛ ⎞ ⎛  N |=Y ⎝¬ ¬φ j ⎠ ∨ ⎝¬ M|= X 0 φ j

⎞ ¬φ j ⎠ .

M|= X 1 φ j

Thus Y = Y0 ∪ Y1 such that N |=Y0 ¬





¬φ j

M|= X 0 φ j

and N |=Y1 ¬



¬φ j .

M|= X 1 φ j

By this and the induction hypothesis, player II has a winning strategy in the positions (X 0 , Y0 ), (X 1 , Y1 ). Thus she can play {Y0 , Y1 } and maintain her winning strategy.

6.6 Ehrenfeucht–Fra¨ıss´e game for dependence logic

127

Case (ii) Player I decides that the game should continue from the new position (X (M/xn ), Y (m/xn )). We claim that M |= X (M/xn ) φ ⇒ N |=Y (N /xn ) φ for all φ ∈ Fmlm n+1 . From this the induction hypothesis would imply that II has a winning strategy in the position (X (M/xn ), Y (N /xn )). So let us assume M |= X (M/xn ) φ, where φ ∈ Fmlm n+1 . By definition, M |= X ¬∃xn ¬φ. Since ¬∃xn ¬φ ∈ Fmlm+1 , (4)m+1 gives N |=Y ¬∃xn ¬φ and n N |=Y (N /xn ) φ follows. Case (iii) Player I chooses a function F : X → M. Let φi , i < M be a complete list (up to logical equivalence) of formulas in Fmlm n+1 . Now  M |= X ∃xn ¬ ¬φi . M|= X (F/xn ) φi

Note that   qr ∃xn ¬

 ¬φ j

M|= X (F/xn ) φ j

 = qr ¬ 



 ¬φ j

+1

M|= X (F/xn ) φ j

 qr(¬¬φ j ) + 1 M|= X (F/xn ) φ j   = max qr(φ j ) + 1 ≤ m + 1,

=

max

M|= X (F/xn ) φ j

and hence, by (4)m+1 , N |=Y ∃xn ¬



¬φi .

M|= X (F/xn ) φi

Thus there is a function G : Y → N such that  N |=Y (G/xn ) ¬ ¬φi . M|= X (F/xn ) φi

The game continues from position (X (F/xn ), Y (G/xn )). Given that now M |= X (F/xn ) φ ⇒ N |=Y (G/xn ) φ for all φ ∈ Fmlm n+1 , the induction hypothesis implies that II has a  winning strategy in position (X (F/xn ), Y (G/xn )).

128

Model theory

Corollary 6.45 Suppose M and N are models of the same vocabulary. Then the following are equivalent. (1) M D N . (2) For all natural numbers n, player II has a winning strategy in the game EFn (M, N ). Corollary 6.46 Suppose M and N are models of the same vocabulary. Then the following are equivalent. (1) M ≡D N . (2) For all natural numbers n, player II has a winning strategy both in the game EFn (M, N ) and in the game EFn (N , M). The two games EFn (M, N ) and EFn (N , M) can be put together into one game simply by making the moves of the former symmetric with respect to M and N . Then player II has a winning strategy in this new game if and only if M ≡nD N . Instead of a game, we could have used a notion of a back-and-forth sequence. The Ehrenfeucht–Fra¨ıss´e game can be used to prove non-expressibility results for D, but we do not yet have examples where a more direct proof using compactness, interpolation, and L¨owenheim–Skolem theorems would not be simpler. Proposition 6.47 There are countable models M and N such that M D N , but N ≡D M. Proof Let M be the standard model of arithmetic. Let n , n ∈ ω, be the list of all 11 -sentences true in M. Suppose n = ∃R1n . . . ∃Rknn φn . Let M∗ be an expansion of M in which each φn is true. Let N ∗ be a countable non-standard elementary extension of M∗ . Let N be the reduct of N ∗ to the language of arithmetic. By construction, M D N . On the other hand, N D M as non well-foundedness of the integers in N can be expressed by a sentence of D.  Proposition 6.48 Suppose K is a model class5 and n is a natural number. Then the following are equivalent: 5

A model class is a class of models, closed under isomorphism, of the same vocabulary.

6.6 Ehrenfeucht–Fra¨ıss´e game for dependence logic

129

Table 6.5. X x0

x1

x2

0 1 2

2 0 1

1 0 1

(1) K is definable in dependence logic by a sentence in Fmln0 ; (2) K is closed under the relation nD . Proof Suppose K is the class of models of φ ∈ Fmln0 . If M |= φ and M nD N , then, by definition, N |= φ. Conversely, suppose K is closed under nD . Let  φM = ¬ {¬φ : φ ∈ Fmln0 , M |= φ}, where the conjunction is taken over a finite set which covers all such φ up to logical equivalence. Let θ be the disjunction of all φM , where M ∈ K . Again we take the disjunction over a finite set up to logical equivalence. We show that K is the class of models of θ . If M ∈ K then M |= φM , whence M |= θ . On the other hand suppose M |= φN for some N ∈ K . Now N nD M, for if N |= φ and φ ∈ Fmln∅ , then φ is logically equivalent with one of the conjuncts of φN , whence M |= φ. As K is closed under nD , we have M ∈ K .  Corollary 6.49 Suppose K is a model class. Then the following are equivalent: (1) K is definable in dependence logic. (2) There is a natural number n such that K is closed under the relation nD . Corollary 6.49 also gives a characterization of 11 -definability in second order logic. No assumptions about cardinalities are involved, so if we restrict ourselves to finite models we get a characterization of NP-definability. Exercise 6.54 Suppose L = ∅ and M is an L-structure such that M = {0, 1, 2}. Consider the team in Table 6.5. List the formulas φ in Fml03 such that X is of type φ in M.

Model theory

130

Table 6.6. X x0

x1

x2

30 1 2 30

30 1 2 30

10 0 1 1

Table 6.7. X x0

x1

x2

0 1 2 3 3 3

0 2 4 8 16 32

0 0 1 1 1 1

Exercise 6.55 Suppose L = ∅ and M is an L-structure such that M = N. Consider the team in Table 6.6. List the formulas φ in Fml03 such that X is of type φ in M. Exercise 6.56 Suppose L = {P} and M is an L-structure such that M = N and P M = {2n : n ∈ N}. Consider the team in Table 6.7. List the formulas φ in Fml03 such that X is of type φ in M. Exercise 6.57 Suppose L = ∅ and M and N are L-structures such that M = {0, 1, 2} and N = {a, b, c, d}, where a, b, c, and d are distinct. The game EF3 (M, N ) ends in the position shown in Table 6.8. Who won the game? Exercise 6.58 Suppose L = ∅ and M and N are L-structures such that M = {0, 1, 2} and N = {a, b, c, d}, where a, b, c, and d are distinct. The game EF3 (M, N ) ends in the position shown in Table 6.9. Who won the game?

6.6 Ehrenfeucht–Fra¨ıss´e game for dependence logic

131

Table 6.8. X

Y

x0

x1

x2

x0

x1

x2

0 1 2

2 0 1

2 0 1

a b c d

c b a d

b c a d

Table 6.9. X

Y

x0

x1

x2

x0

x1

x2

0 1 2

1 1 1

1 0 0

a b c d

c c c c

b b b a

Table 6.10. X

Y

x0

x1

x0

x1

0 1 2 3

2 3 3 2

a b c d

b c d a

Exercise 6.59 Suppose L = ∅ and M and N are L-structures such that M = {0, 1, 2, 3} and N = {a, b, c, d}, where a, b, c, and d are distinct. The game EF3 (M, N ) is in the position shown in Table 6.10. Can you spot a good splitting move for player I? Exercise 6.60 Suppose L = {P} and M and N are L-structures such that M = {0, 1, 2}, P M = {0}, N = {a, b, c}, and P N = ∅, where a, b, and c are distinct. The game EF3 (M, N ) is in the position shown in Table 6.11. Can you spot a good supplementing move for player I?

Model theory

132

Table 6.11. X

Y

x0

x1

x0

x1

0 1 2

2 1 1

a b c

b b a

Exercise 6.61 Let L = ∅. Show that there is no sentence φ of vocabulary L in Fml20 such that M |= φ if and only if M is infinite. Exercise 6.62 Let L = ∅. Show that there is no sentence φ of vocabulary L in Fml20 such that for finite models M we have M |= φ if and only if |M| is even. Exercise 6.63 Show that there is a countable model which is not D-equivalent to any uncountable models. Exercise 6.64 Show that (R, N) D (Q, N). Exercise 6.65 Show that (ω + ω∗ + ω, <) D (ω, <). Exercise 6.66 Show that if M D N and N is a connected graph, then so is M. Exercise 6.67 Give three models M, M , and M such that M D M and M D M , but M D M and M D M . Exercise 6.68 Give models Mn such that for all n ∈ N we have Mn+1 D Mn , but Mn D Mn+1 . Exercise 6.69 Let Part(M, N ) be the set of pairs (X, Y ), where, for some n, both X ⊆ M n and Y ⊆ N n , and, moreover, Y is of the type of any atomic, negated atomic formula, or atomic dependence formulas, φ(x0 , . . . , xn−1 ) in N and X is in M. Suppose M and N are L-structures. A dependence backand-forth set from M to N is any non-empty set P ⊆ Part(M, N ) such that for all (X, Y ) ∈ P: (i) for any X 1 and X 2 such that X = X 1 ∪ X 2 there are Y1 and Y2 such that Y = Y1 ∪ Y2 and both (X 1 , Y1 ) ∈ P and (X 2 , Y2 ) ∈ P; (ii) (X (M/xn ), Y (N /xn )) ∈ P;

6.6 Ehrenfeucht–Fra¨ıss´e game for dependence logic

133

(iii) for any mapping F : X → M there is a mapping G : Y → N such that (X (F/xn ), Y (G/xn )) ∈ P. The structure M is said to be partially dependence isomorphic to N , in symbols M D N , if there is a back-and-forth set from M to N . Show that M D N implies M D N . Exercise 6.70 This continues Exercise 6.69. Show that if M ∼ = N , then M D N . Exercise 6.71 This continues Exercise 6.69. Show that if M D M and M D M , then M D M . Exercise 6.72 This continues Exercise 6.69. Give dense linear orders without end points M and M such that M D M .

7 Complexity

7.1 Decision and other problems The basic problem this chapter considers is how difficult is it to decide some basic questions concerning the relation M |= φ, when M is a structure and φ is a D-sentence? Particular questions we will study are as follows: Decision Problem. Is φ valid, i.e. does M |= φ hold for all M? See Fig. 7.1. Non-validity Problem. Is φ nonvalid, i.e. does φ avoid some model, i.e. does M |= φ fail for some model M? Consistency Problem. Is φ consistent, i.e. does φ have a model, i.e. does M |= φ hold for some M? Inconsistency Problem. Is φ inconsistent, i.e. does φ avoid all models, i.e. is M |= φ true for no model M at all? Obviously such questions depend on the vocabulary. In a unary vocabulary it may be easier to answer some of the above questions. On the other hand, if at least one binary predicate is allowed, then the questions are as hard as for any other vocabulary, as there are coding techniques that allow us to code bigger vocabularies into one binary predicate. We can ask the same questions about models of particular theories such as like groups, linear orders, fields, graphs, and so on. Furthermore, we can ask these questions in the framework of finite models. By definition, φ is non-valid if and only if φ is not valid, φ is inconsistent if and only if φ is not consistent. So it suffices to concentrate on the Decision Problem and the Consistency Problem. In first order logic we have the further equivalence φ is consistent if and only if ¬φ is not valid. 134

7.2 Some set theory

135

Table 7.1. First order logic Problem

Complexity

Decision Problem Non-validity Problem Consistency Problem Inconsistency Problem

10 01 01 10

Yes Is φ valid? No

Fig. 7.1. Machine model of complexity.

So, if we crack the Decision Problem for first order logic, everything else follows. Indeed, the G¨odel Completeness Theorem tells us that a first order sentence is valid if and only if it is provable. Hence the Decision Problem of first order logic is 10 (i.e. recursively enumerable); see Fig 7.2. We obtain Table 7.1. The Consistency Problem for dependence logic can be reduced to that of first order logic by the following equivalence: φ is consistent if and only if τ1,φ is consistent. So we obtain Table 7.2. In Section 7.2 we will replace the two question marks with a complexity class.

7.2 Some set theory The complexity of the Decision Problem and of the Non-validity Problem of dependence logic is so great that we have to move from complexity measures

Complexity

136

Table 7.2. Dependence logic Problem

Complexity

Decision Problem Nonvalidity Problem Consistency Problem Inconsistency Problem

? ? 01 10

Π01

Σ01

recursive

Fig. 7.2. Degrees of decidability.

on the integers to complexity measures in set theory. With this in mind, we recall some elementary concepts from set theory. A set a is transitive if c ∈ b and b ∈ a imply c ∈ a for all a and b. The transitive closure T C(a) of a set a is the intersection of all transitive supersets of a, or, in other words, a ∪ (∪a) ∪ (∪ ∪ a) . . . Intuitively, T C(a) consists of elements of a, elements of elements of a, elements of elements of elements of a, etc. We define Hκ = {a : |T C(a)| < κ}. A priori it is not evident that Hκ is a set. However, this can be easily proved with another useful concept from set theory, namely the concept of rank. The rank rk(a) of a set a is defined recursively as follows: rk(a) = sup{rk(b) + 1 : b ∈ a}. Recall the definition of the cumulative hierarchy in Eq. (4.1); see Fig. 7.3. Now, Vα = {x : rk(x) < α} and we can prove that Hκ is indeed a set. Lemma 7.1 For all infinite cardinals κ we have Hκ ⊆ Vκ .

7.2 Some set theory

137

Vν

Vα+1 Vα

Fig. 7.3. The V -hierarchy of sets.

Proof Suppose x ∈ Hκ , i.e. |T C(x)| < κ. We claim that rk(x) < κ, i.e. | rk(x)| < κ. It suffices to show that | rk(x)| ≤ |T C(x)|. This follows by the Axiom of Choice, if we show that there is, for all x, an onto function from T C(x) onto rk(x). This function is in fact the function z → rk(z). Suppose the claim that rk T C(y) maps T C(y) onto rk(y) holds for all y ∈ x. We show that it holds for x. To this end, suppose α < rk(x). By definition, α ≤ rk(y) for some y ∈ x. If α < rk(y), then, by the induction hypothesis, there is z ∈ T C(y) such that rk(z) = α. Now z ∈ T C(x), so we are done. The other case is that α = rk(y). Since y ∈ T C(x), we are done again.  The converse of Lemma 7.1 is certainly not true in general. For example, the set Vℵ1 has sets such as P(ω) which cannot be in Hℵ1 . However, let us define the Beth numbers as follows: 0 = ω, α+1 = 2α , and ν = limα<ν α for limit ν. For any κ there is λ ≥ κ such that λ = λ , as follows. Let κ0 = κ, κn+1 = κn and λ = limn<ω κn . Then λ = λ . It is easy to see by induction on α that |Vω+α | = α . Lemma 7.2 If κ = κ , then Hκ = Vκ . Proof The claim Hκ ⊆ Vκ follows from Lemma 7.1. On the other hand, if x ∈ Vκ , say x ∈ Vα , where α = ω + α < κ, then |T C(x)| ≤ |Vα | = α < κ = κ, so x ∈ Hκ .  We now recall an important hierarchy in set theory. Definition 7.3 (ref. [30]) The Levy Hierarchy of formulas of set theory is obtained as follows. The 0 -formulas, which are also called 0 -formulas, are all formulas in the vocabulary {∈} obtained from atomic formulas by the

Complexity

138

Σ2

Π2

Σ1

Π1

Fig. 7.4. Levy Hierarchy of formulas.

operations ¬, ∨, ∧, and the following bounded quantifiers: ∃x0 (x0 ∈ x1 ∧ φ) and ∀x0 (x0 ∈ x1 → φ). The n+1 -formulas are obtained from n -formulas by existential quantification (see Fig 7.4). The n+1 -formulas are obtained from n -formulas by existential quantification. A basic property of the 1 -formulas is captured by the following lemma. Lemma 7.4 For any uncountable cardinal κ, a1 , . . . , an ∈ Hκ and 1 -formula φ(x1 , . . . , xn ): (Hκ , ) |= φ(a1 , . . . , an ) if and only if φ(a1 , . . . , an ). Proof The “only if” part is very easy (see Exercise 7.2). For the more difficult direction suppose φ(x1 , . . . , xn ) is of the form ∃x0 ψ(x0 , x1 , . . . , xn ), where ψ(x0 , x1 , . . . , xn ) is 0 and there is an a0 such that φ(a0 , a1 , . . . , an ). Let α be large enough that a0 , . . . an ∈ Vα . Then (Vα , ) |= ψ(a0 , . . . , an ) by Exercise 7.2. Thus (Vα , ) |= φ(a1 , . . . , an ). Let M be an elementary submodel of the model (Vα , ) such that T C({a1 , . . . , an }) ⊆ M and |M| < κ. By Mostowski’s Collapsing Lemma (see Exercise 7.3) there is a transitive model (N , ) and an isomorphism π : (N , ) ∼ = M such that a1 , . . . , an ∈ N and π(ai ) = ai for each i. Thus (N , ) |= φ(a1 , . . . , an ). Since x ∈ N implies

7.2 Some set theory

139

Table 7.3. Dependence logic Problem Decision Problem Nonvalidity Problem Consistency Problem Inconsistency Problem

Complexity 2 2 01 10

T C(x) ⊆ N , and |N | < κ, we have N ⊆ Hκ , and hence again, by absoluteness, (Hκ , ) |= φ(a1 , . . . , an ).  An intuitive picture of a 1 -statement ∃x0 φ(a, x0 ), where φ is 0 , is that we search through the universe for an element b such that the relatively simple statement φ(a, b) becomes true. By Lemma 7.4 we need only look near where a is. This means that satisfying a 1 -sentence is, from a set theoretic point of view, not very complex, although it may still be at least as difficult as checking whether a recursive binary relation on N is well-founded (which is a 11 -complete problem). In contrast, to check whether a 2 -sentence ∃x0 ∀x1 φ(a, x0 , x1 ) is true, one has to search through the whole universe for a b such that ∀x1 φ(a, b, x1 ) is true. Now, we cannot limit ourselves to search close to a as we may have to look close to b, too. So, in the end, we have to go through the whole universe in search of b. This means that checking the truth of a 2 -sentence is of extremely high complexity. Any of the following statements can be expressed as the truth of a 2 -sentence: (i) (ii) (iii) (iv) (v)

the Continuum Hypothesis, i.e. 2ℵ0 = ℵ1 ; the failure of the Continuum Hypothesis, i.e. 2ℵ0 = ℵ1 ; V = L; there is an inaccessible cardinal; there is a measurable cardinal.

We now have the notions at hand for filling in the complexity table (Table 7.3) for dependence logic, even if we have not yet proved anything. The proof is given in Section 7.3. Exercise 7.1 Show that the following predicates of set theory can be defined with a 0 -formula:

Complexity

140

(i) (ii) (iii) (iv) (v)

x x x x x

= y ∪ z; = {y, z}; is transitive; is an ordinal (i.e. a transitive set of transitive sets); is a function y → z.

Exercise 7.2 (Absoluteness of 0 ) For any transitive M, a1 , . . . , an ∈ M and 0 -formula φ(x1 , . . . , xn ): (M, ) |= φ(a1 , . . . , an ) if and only if φ(a1 , . . . , an ). Exercise 7.3 (Mostowski’s Collapsing Lemma) Suppose (M, E) is a wellfounded model of the Axiom of Extensionality: ∀x0 ∀x1 (∀x2 (x2 ∈ x0 ↔ x2 ∈ x1 ) → x0 = x1 ). Show that the equation π (x) = {π (y) : y E x} defines an isomorphism between (M, E) and (N , ∈), where N is a transitive set. Show also that if E = ∈, then π(x) = x for every x ∈ M which is transitive.

7.3 2 -completeness in set theory Let us now go a little deeper into details. We identify problems with sets X ⊆ N. The problem X in such a case is really the problem of deciding whether a given n is in X or not. A problem P ⊆ N is called n -definable if there is a n -formula φ(x0 ) of set theory such that n ∈ P ⇐⇒ φ(n). A problem P ⊆ N is called n -complete if the problem itself is n -definable and, moreover, for every n -definable set X ⊆ N there is a recursive function f : N → N such that n ∈ X ⇐⇒ f (n) ∈ P. The concepts of a n -definable set and a n -complete set are defined analogously. Theorem 7.5 The Decision Problem of dependence logic is 2 -complete in set theory. Proof Without loss of generality, we consider the decision problem in the context of a vocabulary consisting of just one binary predicate. Let us first observe that the predicate x = P(y) is 1 -definable: x = P(y) ⇐⇒ ∀z(z ∈ x ↔ ∀u ∈ z(u ∈ y)).

7.3 2 -completeness in set theory

141

Let On(x) be the 0 -predicate “x is an ordinal,” i.e. x is a transitive set of transitive sets. We can 1 -define the property R(x) of x of being equal to some Vα , where α is a limit ordinal. Let Str(x) be the first order formula in the language of set theory which says that x is a structure of the vocabulary containing just one binary predicate symbol. If φ is a sentence of D, let Satφ (x) be the first order formula in the language of set theory which says “Str(x) and φ is true in the structure x.” Let Relsatφ (x, y) be the first order formula in the language of set theory which says “x ∈ y and if Str(x), then Satφ (x) is true when relativized to the set y.” Thus |= φ if and only if ∀x(Str(x) → Satφ (x)). Note that for limit α and a ∈ Vα : Satφ (a) ⇐⇒ (Vα |= Satφ (a)). Thus a sentence φ of D is valid if and only if ∀x(R(x) → ∀y ∈ xRelsatφ (y, x)). We have proved that the Decision Problem of D is 2 -definable. Suppose then that A is an arbitrary 2 -definable set of integers. Let ∀x∃yψ(n, x, y) be the 2 -definition. Let φn be the first order sentence ∀x∃yψ(n, x, y), where n is a defined term. We claim n ∈ A ⇐⇒|= θ ∨ φn , where θ is a sentence of D which is true in every model except the models (Vκ , ∈), κ = κ . Suppose first n ∈ A, i.e. ∀x∃yψ(n, x, y). Suppose (Vκ , ∈) is a given model in which θ is not true. We prove Vκ |= ∀x∃yψ(n, x, y). Suppose a ∈ Vκ . By the above lemmas, there is b ∈ Vκ such that ψ(n, a, b). We have proved |= θ ∨ φn . Conversely, suppose |= θ ∨ φn . To prove ∀x∃yψ(n, x, y), let a be given. Let κ be an infinite cardinal such that κ = κ and a ∈ Hκ . Then there is b ∈ Hκ with Hκ |= ψ(n, a, b). Now ψ(n, a, b) follows.  Corollary 7.6 The Decision Problem of dependence logic is not 2 -definable in set theory. The fact that the Decision Problem of dependence logic is not nm for any m, n < ω follows easily from this. Moreover, it follows that we cannot in general express “φ is valid,” for φ ∈ D, even by searching through the whole set theoretical universe for a set x such that a universal quantification over the subsets of x would guarantee the validity of φ. In contrast, to check the validity of a first order sentence, one needs only to search through all natural numbers and then perform a finite polynomial calculation on that number. Figures 7.5 and 7.6 illustrate the difference between categories of validity in first order logic and dependence logic.

Complexity

142

True in some models and false in some models

True in all models

False in all models

Fig. 7.5. Categories of validity in first order logic.

Non-determined

True in some models

False in some models

True in all models

False in all models

Fig. 7.6. Categories of validity in dependence logic.

Exercise 7.4 Give a 2 -definition of the property of x being equal to some Vκ , where κ = κ . Exercise 7.5 Show that for limit α and a ∈ Vα : Satφ (a) ⇐⇒ (Vα |= Satφ (a)).

7.3 2 -completeness in set theory

143

Exercise 7.6 Give a sentence of D which is true in every model except (mod ∼ =) the models (Vκ , ∈), κ = κ . Exercise 7.7 Use Theorem 7.5 to prove that the Decision Problem of D is not arithmetical, i.e. not first order definable in (N, +, ·, <, 0, 1). (In fact, the same proof shows that it is not nm -definable for any m, n < ω.) Exercise 7.8 Show that the problem “does M |= φ hold for all countable M?” is not arithmetical and that the problem “does M |= φ hold for some countable M?” is a 01 -property of φ ∈ D. Exercise 7.9 Show that there is a sentence φ of D such that φ avoids some model if and only if there is an inaccessible cardinal.

8 Team logic

The negation ¬ of dependence logic does not satisfy the Law of Excluded Middle and is therefore not the classical Boolean negation. This is clearly manifested by the existence of non-determined sentences φ in D. In such cases, the failure of M |= φ does not imply M |= ¬φ. Hintikka [19] introduced extended independence friendly logic by taking the Boolean closure of his independence friendly logic. We take the further action of making classical negation ∼ one of the logical operations on a par with other propositional operations and quantifiers. This yields an extension TL of the Boolean closure of D. We call the new logic team logic. The basic concept of both team logic TL and dependence logic D is the concept of dependence =(t1 , . . . , tn ). In very simple terms, what happens is that, while we can say “x1 is a function of x0 ” with =(x0 , x1 ) in D, we will be able to say “x1 is not a function of x0 ” with ∼ =(x0 , x1 ) in team logic. While we define team logic we have to restrict ¬. The game theoretic intuition behind ¬φ is that it says something about “the other player.” The introduction of ∼ unfortunately ruins the basic game theoretic intuition, and there is no “other player” anymore. If φ is in D, then ∼ φ has the meaning “II does not have a winning strategy,” but it is not clear what the meaning of ¬ ∼ φ would be. We also change notation by using φ ⊗ ψ (“tensor”) instead of φ ∨ ψ. The reason for this is that by means of ∼ and ∧ we can actually define the classical Boolean disjunction φ ∨ ψ, which really says that the team is of type φ or of type ψ. Likewise, we adopt the notation !xn φ (“shriek”) for the quantifier that in D was denoted by ∀xn φ.

8.1 Preorder of determination Let us consider teams of soccer players as an example. In this case, players are the agents. We consider only the colors of their outfits as relevant 144

8.1 Preorder of determination

145

Table 8.1. Team 1

1 2 3 4 5 6 7

Shirt

Shorts

yellow yellow yellow yellow red red red

white white white white black black black

Table 8.2. Team 2

1 2 3 4 5 6 7

Shirt

Shorts

Socks

yellow yellow yellow red red red red

white white white white white black black

green green green black black green green

features. We imagine a spectator trying to figure out how the different colors are determined. Consider Team 1 of Table 8.1, consisting of seven members. It is clear that in this team both the shirt color depends on the shorts color and vice versa. What about Team 2 of Table 8.2? We can observe that the shirt color depends on shorts color and socks’ color in the sense that it can be computed if both shorts color and socks color are known. Interestingly, it cannot be computed from either shorts or socks alone. It seems correct to say that shirt color is dependent on shorts color, and that shirt color is dependent on socks color. However, if the team was as in Table 8.3, we could compute shirt color from socks color alone. It would seem correct to say that shirt color is, in this case, independent of shorts color, as the only addition that together with the shorts color would determine shirt color, achieves this by itself. Let us finally consider Table 8.4. Now shirt color can be computed alternatively from shorts color alone or from the combination of socks and shoes. We certainly cannot say

Team logic

146

Table 8.3. Team 3

1 2 3 4 5 6 7

Shirt

Shorts

Socks

yellow yellow yellow red red red red

white white white white white black black

green green green black black red red

Table 8.4. Team 4

1 2 3 4 5 6 7

Shirt

Shorts

Socks

Shoes

yellow yellow yellow red red red red

white white white blue blue black black

green green green green green yellow yellow

yellow yellow yellow red red yellow yellow

shirt color is independent of shorts color, even if it can be computed without it. Similarly, shirt color is not independent of socks color, as socks color (together with shoes color) gives away shirt color, but shoes color alone does not. We can say that shirt color depends on shorts color, but not so strongly that it could not be computed without it. Another example is the human genome “team.” Here an agent is the DNA sequence of the set of chromosomes of any one individual human being. Thus there are over 6.5 billion agents from people alive today. A team is any collection of such agents. Features are any of the 25 000 genes that make up the DNA sequence of the individual person. Extra features such as diseases (diabetes, asthma, heart disease, etc.) can be added to genome teams to study how faulty genes are associated with particular diseases. Then a team would be a patient database with fields for certain genes and certain diseases. In anticipation of the discussion in Section 8.2, of the different kinds of dependencies, we may ask the following kinds of questions.

8.1 Preorder of determination

147

Table 8.5. A team

1 2 3 4 5 ... k

x1

x2

...

xn

m 11 m 21 m 31 m 41 m 51 ... m k1

m 12 m 22 m 32 m 42 m 52 ... m k2

... ... ... ... ... ... ...

m 1n m 2n m 3n m 4n m 5n ... m kn

(i) Does a certain gene or gene combination (significantly) determine a given hereditary disease in the sense that a patient with (a fault in) those genes has a high risk of the disease? (ii) Is a disease totally dependent on a gene in the sense that every gene combination that (significantly) determines the disease contains that particular gene? (iii) Is a gene (merely) dependent on a gene in the sense that the disease is (significantly) determined by some gene combination with the gene but not without? (iv) Is a disease totally independent of a gene in the sense that no gene combination that (significantly) determines the disease contains that particular gene? (v) Is a gene (merely) independent of a gene in the sense that some gene combinations (significantly) determine the disease without containing that particular gene? Let X be a team, i.e. a set of assignments (agents) of some finitely many variables {x1 , . . . , xn } into a domain M, as in Table 8.5. Equivalently, X could be a database with fields {x1 , . . . , xn } and values of the fields from a fixed domain M. There is an obvious partial order in the powerset of {x1 , . . . , xn }, namely the set theoretical subset relation ⊆. We now define a new relation, called the preorder of determination: V ≤W

W determines V , i.e. features in V can be determined if the values of the features in W are known. In symbols, ∀s, s ∈ X ((∀y ∈ W (s(y) = s (y)) → (∀x ∈ V (s(x) = s (x))))).

Equivalently, {w1 , . . . , wn } determines V , if, for all y ∈ V , there is a function f y such that for all s in X : s(y) = f y (s(w1 ), . . . , s(wn )).

Team logic

148

It is evident from the definition that the preorder of determination is weaker than the partial order of inclusion in the sense that a set obviously determines every subset of itself. It is more interesting that sometimes a set determines a set disjoint from itself, and a singleton set may determine another singleton set. Some sets may be determined by the empty set (in that case the feature has to have a constant value). Every set is certainly determined by the whole universe. The Armstrong Axioms of functional dependence (see ref. [1]) state exactly the following lemma. Lemma 8.1 (i) V ≤ W is a preorder, i.e. reflexive and transitive. (ii) If V ⊆ W , then V ≤ W . (iii) If V ≤ W and U is arbitrary, then V ∪ U ≤ W ∪ U . Moreover, (iv) If V ≤ W , then there is a minimal U ⊆ W such that V ≤ U .

8.2 Dependence and independence Now we can define two versions of dependence, by using the preorder of determination. Suppose W ∩ V = ∅. We define the following. V is dependent on W

There is some minimal U ≥ V such that U ∩ V = ∅ and W ⊆ U .

V is totally dependent on W

For every U ≥ V such that U ∩ V = ∅, we have W ⊆ U .

V is independent of W

There is some U ≥ V such that U ∩ V = ∅ and W ∩ U = ∅.

V is totally independent of W

For every minimal U ≥ V such that U ∩ V = ∅, we have W ∩ U = ∅.

V is non-determined

There is no U ≥ V such that U ∩ V = ∅. In the opposite case, V is called determined.

Note that the above concepts are defined with respect to the features that are present in the setup. Indeed, it seems meaningless to define what independence means in a setup where any new feature can be introduced. The presence of new features can change independence to dependence, as in Team 2 of Table 8.2,

8.2 Dependence and independence

149

Table 8.6. Team for Exercise 8.3

s1 s2 s3

x1

x2

x3

0 0 1

1 0 1

0 1 1

where shirt color is independent of both shorts and socks if either one of them is missing, but is dependent on either if both are present. In the following, we present some immediate relationships between the introduced concepts of dependence and independence. Lemma 8.2 (i) Every V is totally dependent on and totally independent of ∅. (ii) If V is (totally) dependent on W , then V is (totally) dependent on every subset of W . (iii) If V is dependent on W , it can still be also independent of W , but not totally, unless W = ∅. (iv) V is independent of {x} if and only if V is not totally dependent on {x}. (v) V is totally independent of {x} if and only if V is not dependent on {x}. Exercise 8.1 Prove Lemma 8.1. Exercise 8.2 Prove Lemma 8.2. Exercise 8.3 Consider the team given in Table 8.6. Which W ⊆ {x2 , x3 } is {x1 } (a) dependent on, (b) totally dependent on, (c) independent of, (d) totally independent of? Exercise 8.4 Consider the team given in Table 8.7. Which subsets W of {x2 , x3 , x4 , x5 } is {x1 } (a) dependent on, (b) totally dependent on, (c) independent of, (d) totally independent of? Exercise 8.5 Consider the team given in Table 8.8. Find a value for a in natural numbers such that (a) {x3 , x4 } determines {x1 }, (b) {x3 , x4 } does not determine {x1 }, (c) {x2 , x4 } determines {x1 }, (d) {x2 , x4 } does not determine {x1 }? Exercise 8.6 Consider the team given in Table 8.9. Fill in values for x3 and x4 from the set {0, 1} such that {x4 } is (a) dependent but not totally dependent

Team logic

150

Table 8.7. Team for Exercise 8.4

s1 s2 s3 s4 s5 s6

x1

x2

x3

x4

x5

2 2 8 0 0 2

3 2 0 3 0 4

0 4 4 4 0 0

1 0 0 1 0 1

0 3 0 1 0 1

Table 8.8. Team for Exercise 8.5

s1 s2 s3 s4 s5

x1

x2

x3

x4

15 1 6 4 0

15 1 0 3 0

3 1 2 2 0

6 1 a 2 5

Table 8.9. Team for Exercise 8.6

s1 s2 s3

x1

x2

0 0 1

0 1 1

x3

x4

on, (b) totally dependent on, (c) independent but not totally independent of, (d) totally independent of {x1 }.

8.3 Formulas of team logic In this section we give the syntax and semantics of team logic and indicate some basic principles.

8.3 Formulas of team logic

151

Table 8.10. Atomic

Name

t1 = tn ¬t1 = tn Rt1 . . . tn ¬Rt1 . . . tn = (t1 , . . . , tn ) ¬ = (t1 , . . . , tn )

equation dual equation relation dual relation dependence dual dependence

Table 8.11. Compound

Name

φ⊗ψ φ∧ψ ∼φ ∃xn φ !xn φ

tensor conjunction negation existential shriek

Definition 8.3 Suppose L is a vocabulary. The formulas of team logic TL are of the atomic form, as in Table 8.10, or compound form, as in Table 8.11. The semantics of team logic is defined as follows. Definition 8.4 (TL1) M |= X t1 = t2 if and only if for all s ∈ X we have t1M s = t2M s. (TL2) M |= X ¬t1 = t2 if and only if for all s ∈ X we have t1M s = t2M s. (TL3) M |= X =(t1 , . . . , tn ) if and only if for all s, s ∈ X such that M M t1M s = t1M s , . . . , tn−1 s = tn−1 s ,

we have tnM s = tnM s . (TL4) M |= X ¬ =(t1 , . . . , tn ) if and only if X = ∅. (TL5) M |= X Rt1 . . . tn if and only if for all s ∈ X we have (t1M s, . . . , tnM s) ∈ R M . (TL6) M |= X ¬Rt1 . . . tn if and only if for all s ∈ X we have (t1M s, . . . , tnM s) ∈ / RM.

Team logic

152 (TL7) M |= X M |=Y (TL8) M |= X (TL9) M |= X (TL10) M |= X (TL11) M |= X

φ ⊗ ψ if and only if X = Y ∪ Z such that dom(Y ) = dom(Z ), φ, and M |= Z ψ. φ ∧ ψ if and only if M |= X φ and M |= X ψ. ∃xn φ if and only if M |= X (F/xn ) φ for some F : X → M. !xn φ if and only if M |= X (M/xn ) φ ∼ φ if and only if M |= X φ.

We can easily define a translation φ → φ ∗ of dependence logic into team logic, but we have to assume the formula φ of dependence logic is in negation normal form: (t = t )∗ (¬t = t )∗ (Rt1 . . . tn )∗ (¬Rt1 . . . tn )∗ (=(t1 , . . . , tn ))∗ (¬ =(t1 , . . . , tn ))∗ (φ ∨ ψ)∗ (φ ∧ ψ)∗ (∃xn φ)∗ (∀xn φ)∗

= = = = = = = = = =

t = t , ¬t = t , Rt1 . . . tn , ¬Rt1 . . . tn , =(t1 , . . . , tn ), ¬ =(t1 , . . . , tn ), φ∗ ⊗ ψ ∗, φ∗ ∧ ψ ∗, ∃xn φ ∗ , !xn φ ∗ .

It is an immediate consequence of the definitions that for all M, all φ, and all X we have M |= X φ in D if and only if M |= X φ ∗ in TL. So we may consider D a fragment of TL, and TL an extension of D obtained by adding classical negation. Logical consequence φ ⇒ ψ and logical equivalence φ ⇔ ψ are defined similarly as for dependence logic. Lemma 8.5 demonstrates that even though ! xn φ for φ ∈ D acts like what is denoted by ∀xn φ in dependence logic, it is, in the presence of ∼, not at all like our familiar universal quantifier, as it commutes with negation. Lemma 8.5 ∼ !xn φ ⇔ !xn ∼ φ. Proof M |= X ∼ !xn φ

if and only if if and only if if and only if if and only if

M |= X !xn φ, M |= X (M/xn ) φ, M |= X (M/xn ) ∼ φ, M |= X !xn ∼ φ.



8.3 Formulas of team logic

153

Table 8.12. The intuition behind the logical operations φ∨ψ φ∧ψ φ ⊗ψ φ ⊕ψ ∼φ  ⊥ 1 0

the team is either of type φ or of type ψ (or both) the team is both of type φ and of type ψ the team divides between type φ and type ψ every division of the team yields type φ or type ψ the team is not of type φ any team no team any non-empty team only the empty team

We adopt the following abbreviations (see Table 8.12):1 φ∨ψ φ⊕ψ φ→ψ φ −◦ ψ ∀xn φ

= ∼(∼ φ ∧ ∼ ψ) disjunction; = ∼(∼ φ ⊗ ∼ ψ) sum; = ∼φ ∨ ψ implication; = ∼φ ⊕ ψ linear implication; = ∼ ∃xn ∼ φ universal quantifier.

Thus, M |= X φ ∨ ψ if and only if M |= X φ or M |= X ψ. We have recovered the classical disjunction with the properties φ∨ψ ⇔ ψ ∨ φ, φ ∨ (ψ ∨ θ ) ⇔ (φ ∨ ψ) ∨ θ, φ∨φ ⇔ φ. Note that M |= X φ ⊕ ψ if and only if whenever X = Y ∪ Z , then M |=Y φ or M |= Z ψ. Thus a team of type φ ⊕ ψ has in a sense φ or ψ everywhere. We have φ ⊕ψ ⇔ ψ ⊕ φ, φ ⊕ (ψ ⊕ θ) ⇔ (φ ⊕ ψ) ⊕ θ, but φ ⊕ φ  φ. Note also that M |= X ∀xn φ if and only if M |= X (F/xn ) φ holds for every F : X → M. 1

Note that the symbols ∨ and ∀ have different meanings in D and in TL.

Team logic

154

Table 8.13.

s1 s2 s3 s4

x0

x1

3 2 1 9

8 5 4 1

Table 8.14. Dependence values in D Sentence

Teams of that type

 ⊥ 1 0

∅, {∅} (none) {∅} ∅

Thus a team of type ∀xn φ has type φ whatever we put as values of xn . A team of type ∀x2 ((x0 + x1 ) + x2 = x0 + (x1 + x2 )) but not of type ∀x2 (∼ =(x0 , x1 , x2 )) in the model (N, +) is shown in Table 8.13. We have ∀xn (φ ∧ ψ) ⇔ (∀xn φ ∧ ∀xn ψ), ∀xn ∀xm φ ⇔ ∀xm ∀xn φ, but in general (see Lemma 8.5) ∀xn φ  ! xn φ. Definition 8.6 The dependence values are the following special sentences of team logic (see Fig. 8.1 and Table 8.14):  ⊥ 0 1

= = = =

=(), ∼ =(), ¬ =(), ∼ ¬ =().

What about ¬ ∼ =()? This is not a sentence of team logic at all!

8.3 Formulas of team logic

1

155

0 ⊥

Fig. 8.1. The diamond of dependence values.

Example 8.7 Here are some trivial relations between the dependence values: (i) (ii) (iii) (iv) (v)

⊥ ⇒ 0 ⇒ ; ⊥ ⇒ 1 ⇒ ; 1 = ∼ 0, ⊥ = ∼ ; 0 ⇔ ∼ 1,  ⇔ ∼ ⊥; 0 = ¬.

The equation X = X ∪ X yields (φ ∧ ψ) ⇒ (φ ⊗ ψ), (φ ⊕ ψ) ⇒ (φ ∨ ψ). The equation X = X ∪ ∅ yields φ ⇒ (φ ⊗ 0), (φ ⊕ ⊥) ⇒ ∼ φ. The logic TL is much stronger than D. Let us immediately note the failure of compactness. Proposition 8.8 The logic TL does not satisfy the Compactness Theorem. Proof Let φn be the sentence ∀x0 . . . ∀xn ∃xn+1 (¬x0 = xn+1 ∧ . . . ∧ ¬xn = xn+1 ). Then, any finite subset of T = {φn : n ∈ N} ∪ {∼(∞ )∗ } has a model, but T itself does not have a model. 

156

Team logic

Proposition 8.9 The logic TL does not satisfy the L¨owenheim–Skolem Theorem. There is a sentence φ of team logic such that φ has an infinite model, but φ has no uncountable models. There is also a sentence ψ of team logic such that ψ has an uncountable model, but no countable models. Proof Recall Lemma 4.1 Let φ be the conjunction of P − and ∼ ∃x5 ∃x4 ! x0 ∃x1 ! x2 ∃x3 (=(x2 , x3 ) ∧ x4 < x5 ∧ (((x0 = x2 ∧ x0 < x5 ) ∧ (x1 = x3 ∧ x1 < x4 )) ∧ ((¬x0 = x2 ⊗ ¬x0 < x5 ) ∧ (¬x1 = x3 ⊗ ¬x1 < x4 )))). Then φ has the infinite model (N, +, ·, 0, 1, <). But since every model of φ is isomorphic to (N, +, ·, 0, 1, <), it cannot have any uncountable models. For ψ, recall the sentence cmpl from Section 4.3. Let ψ be the conjunction of the axioms of dense linear order and ∼ cmpl . Now ψ has models, e.g. (R, <), but every model is a dense complete order and is therefore uncountable.  It follows that there cannot be any translation of team logic into 11 , as such a translation would yield both the Compactness Theorem and the L¨owenheim– Skolem Theorem as a consequence. With a translation to 11 ruled out, it is difficult to imagine what a game theoretical semantics of team logic would look like. Note that there cannot be a truth definition for TL in TL. Suppose τ ∗ (x0 ) is in TL and M |= φ is equivalent to M |= τ ∗ (φ) in all Peano models M. By using the formula ∼ τ ∗ (x0 ), we can reprove Tarski’s Undefinability of Truth Theorem 6.20 Despite apparent similarities, team logic and linear logic [11] have very little to do with each other. In linear logic, resources are split into “consumable” parts. In team logic resources are split into “coherent” parts. Exercise 8.7 Show that every formula of TL in which ∼ does not occur is logically equivalent with a sentence of D. Exercise 8.8 Prove Example 8.7. Exercise 8.9 Prove the following equivalences: φ∧ φ∨ ⊗ φ⊕

⇔ ⇔ ⇔ ⇔

φ, , , .

8.3 Formulas of team logic

157

Exercise 8.10 Prove the following equivalences: φ∧⊥ φ∨⊥ φ⊗⊥ ⊥⊕⊥

⇔ ⇔ ⇔ ⇔

⊥, φ, ⊥, ⊥.

Exercise 8.11 Prove the following equivalences: φ∗ ∧ 0 φ∗ ∨ 0 φ⊗0 0⊕0

⇔ ⇔ ⇔ ⇔

φ∗ φ∗ φ, 0.

if φ ∈ D, if φ ∈ D,

Exercise 8.12 Prove the following equivalences: 1 ⊗ 1 ⇔ 1, 1 ⊕ 1 ⇔ 1. Exercise 8.13 Give an example which demonstrates ∀xn φ  ! xn φ. Exercise 8.14 Suppose φ is a formula. Give a formula ψ with the property that a team X is of type ψ if and only if every subset of X is of type φ. Exercise 8.15 Suppose φ is a formula. Give a formula ψ with the property that a team X is of type ψ if and only if every subset of X has a subset which is of type φ. Exercise 8.16 Give a formula φ with the property that a team X is of type φ if and only if every subset of X has a subset which is of type φ, but it is not true that a team X is of type φ if and only if every subset of X is of type φ. Exercise 8.17 Show that ∼ is not definable from the other symbols in team logic, i.e show that the sentence ∼ P, where P is a 0-ary predicate symbol, is not logically equivalent to any sentence of team logic of vocabulary {P} without ∼. Exercise 8.18 Show that =(x1 , . . . , xn ) is definable from the other symbols in team logic and formulas of the form =(t). That is, show that there is a formula φ(x1 , . . . , xn ) in team logic such that =(x1 , . . . , xn ) and φ(x1 , . . . , xn ) are logically equivalent and φ(x1 , . . . , xn ) has no occurrences of atomic formulas of the form =(t1 , . . . , tm ), where m ≥ 2.

Team logic

158

Table 8.15. Atomic

Name

t1 = tn Rt1 . . . tn X nm t1 . . . tm

equation relation second order atomic

Table 8.16. Compound

Name

φ∨ψ φ∧ψ ∼φ ∃xn φ ∃X nm φ

disjunction conjunction negation existential first order existential second order

8.4 From team logic to L 2 One of the main features of dependence logic is its equivalence with the 11 part of second order logic, established in Chapter 6. Since 11 is not closed under classical negation, we cannot do the same for the extension TL of D. Instead we translate team logic into full second order logic L 2 . Definition 8.10 Suppose L is a vocabulary. Second order logic L 2 has, in addition to the (individual) variables xn , another type of variable, namely relation variables X nm . The variables X nm are said to be m-ary, and will range over m-ary relations on the domain. The formulas of L 2 are of the atomic form as in Table 8.15 or compound form as in Table 8.16. The semantics of second order logic for a model M is defined by adopting the concept of a second order assignment for M. These are functions S such that, for all m and n, S(X nm ) ⊆ M m . If A ⊆ M m , then the modified assignment S(A/ X nm ) maps X ij to S(X ij ), if i = m or j = n, and X nm to A. Definition 8.11 r M |= t = t if and only if t M s = t M s; s,S 1 2 1 2 r M |= Rt . . . t if and only if (t M s, . . . , t M s) ∈ R M ; s,S 1 n n 1 r M |= X m t . . . t if and only if (t M s, . . . , t M s) ∈ S(m, n); s,S m n 1 m 1

8.4 From team logic to L 2 r r r r r

M |=s,S M |=s,S M |=s,S M |=s,S M |=s,S

159

φ ∨ ψ if and only if M |=s,S φ or M |=s,S ψ; φ ∧ ψ if and only if M |=s,S φ and M |=s,S ψ; ∼ φ if and only if M |=s,S φ; ∃xn φ if and only if M |=s(a/xn ),S φ for some a ∈ M; ∃X nm φ if and only if M |=s,S(A/ X nm ) φ for some A ⊆ M m .

The definitions of both the syntax and the semantics of L 2 look deceptively similar to those of first order logic. However, L 2 is very different from first order logic. For example, L 2 does not satisfy the Compactness Theorem, nor the L¨owenheim–Skolem Theorem, and its Decision Problem is 2 -complete like that of D. We give in this chapter translations of team logic into L 2 and, in a weaker sense, of L 2 into team logic. Theorem 8.12 We can associate with every formula φ(xi1 , . . . , xin ) of team logic in vocabulary L an L 2 -sentence ηφ (U ), where U is n-ary, such that for all L-structures M and teams X with dom(X ) = {i 1 , . . . , i n } the following are equivalent: (i) M |= X φ; (ii) (M, rel(X )) |= ηφ (U ). Proof Exactly as in the proof of Theorem 6.2. Case (1) Suppose φ(xi1 , . . . , xin ) is t1 = t2 or Rt1 . . . tn . Let ηφ (U ) be given by ∀xi1 . . . ∀xin (U xi1 . . . xin → φ(xi1 , . . . , xin )). Case (2) Suppose φ(xi1 , . . . , xin ) is the dependence formula =(t1 (xi1 , . . . , xin ), . . . , tm (xi1 , . . . , xin )), where i 1 < . . . < i n . We define ηφ (U ) as follows. Subcase (2.1) m = 0; we let ηφ (U ) =  and η¬φ (U ) = ⊥. Subcase (2.2) m = 1; now φ(xi1 , . . . , xin ) is the dependence formula = (t1 (xi1 , . . . , xin )). We let ηφ (U ) be the formula ∀xi1 . . . ∀xin ∀xin +1 . . . ∀xin +n ((U xi1 . . . xin ∧ U xin +1 . . . xin +n ) → t1 (xi1 , . . . , xin ) = t1 (xin +1 , . . . , xin +n )), and we further let η¬φ (U ) be the formula ∀xi1 . . . ∀xin ¬U xi1 . . . xin .

160

Team logic Subcase (2.3) If m > 1 we let ηφ (U ) be the formula ∀xi1 . . . ∀xin ∀xin +1 . . . ∀xin +n ((U xi1 . . . xin ∧ U xin +1 . . . xin +n ∧ t1 (xi1 , . . . , xin ) = t1 (xin +1 , . . . , xin +n ) ∧... tm−1 (xi1 , . . . , xin ) = tm−1 (xin +1 , . . . , xin +n )) → tm (xi1 , . . . , xin ) = tm (xin +1 , . . . , xin +n )),

and we further let η¬φ (U ) be the formula ∀xi1 . . . ∀xin ¬U xi1 . . . xin . Case (3) Suppose φ(xi1 , . . . , xin ) is the disjunction (ψ(x j1 , . . . , x j p ) ⊗ θ (xk1 , . . . , xkq )), where {i 1 , . . . , i n } = { j1 , . . . , j p } ∪ {k1 , . . . , kq }. We let the sentence ηφ (U ) be ∃R∃T (ηψ (R) ∧ ηθ (T ) ∧ ∀xi1 . . . ∀xin (U xi1 . . . xin → (Rx j1 . . . x j p ∨ T xk1 . . . xkq ))). Case (4) Suppose φ(xi1 , . . . , xin ) is the conjunction (ψ(x j1 , . . . , x j p ) ∧ θ(xk1 , . . . , xkq )), where {i 1 , . . . , i n } = { j1 , . . . , j p } ∪ {k1 , . . . , kq }. We let the sentence ηφ (U ) be ηψ (U ) ∧ ηθ (U ). Case (5) φ is ∼ ψ; ηφ (U ) is the formula ∼ ηψ (U ). Case (6) Suppose φ(xi1 , . . . , xin ) is the formula ∃xin+1 ψ(xi1 , . . . , xin+1 ). τ1,φ (U ) is the formula ∃R(τ1,ψ (R) ∧ ∀xi1 . . . ∀xin (U xi1 . . . xin → ∃xin+1 Rxi1 . . . xin+1 )). Case (7) Suppose φ(xi1 , . . . , xin ) is the formula !xin+1 ψ(xi1 , . . . , xin+1 ). τ1,φ (U ) is the formula ∃R(ηψ (R) ∧ ∀xi1 . . . ∀xin+1 (U xi1 . . . xin ↔ Rxi1 . . . xin+1 )).



Corollary 8.13 For every sentence φ of team logic there is an L 2 -sentence ηφ such that for all models M we have M |= φ if and only if M |= ηφ . Proof Let ηφ be the result of replacing in ηφ (U ) every occurrence of the 0-ary relation symbol U by . Now the claim follows from Theorem 8.12.  With the translation φ → ηφ we can consider team logic TL a fragment of second order logic L 2 , even if the origin of team logic in the dependence relation =(t1 , . . . , tn ) is totally different from the origin of second order logic,

8.5 From L 2 to team logic

161

and even if the logical operations of team logic are totally different from those of second order logic. We can draw many immediate conclusions from Corollary 8.13 and from what is known about second order logic (see, e.g., ref. [40]). Corollary 8.14 The Decision Problem of team logic is 2 -complete. The Consistency Problem is 2 -complete. Corollary 8.15 If κ is measurable and φ ∈ TL has a model M of cardinality κ, then φ is true in a submodel N of M of cardinality < κ. If κ is supercompact and φ ∈ TL has a model M of any cardinality, then φ is true in a submodel N of M of cardinality < κ. For the definition of measurable and supercompact cardinals we refer to ref. [25]. Proof This follows from the similar result for L 2 [32].



8.5 From L 2 to team logic We have given a translation of team logic into second order logic. Now we give an implicit translation of second and higher order logic into team logic. The translation is implicit in the sense that it uses new predicates and an extension of the universe. However, the new predicates and the new universe are unique up to isomorphism. Theorem 8.16 Suppose L is a vocabulary and n ∈ N. There is a sentence  of TL in the vocabulary L = L ∪ {P, E} such that for all L-structures M there is N a unique (mod ∼ =) N |=  with (N L)(P ) = M. Moreover, we can associate with every sentence φ of L 2 , with second order variables of any arity ≤ n, in vocabulary L a TL-sentence ξφ in the vocabulary L such that the following are equivalent for all L-structures M: (i) M |= φ; N (ii) N |= ξφ for the unique (mod ∼ =) N such that N |=  and (N L)(P ) = M. Proof To avoid cumbersome notation, we restrict ourselves to second order formulas with only the unary second order variables X n1 . We assume, again for simplicity, that individual variables x0 , x2 , x4 , . . . have not been used in our second order formulas. Clearly this is harmless as one can change bound variables. The idea is that x2n is a new “first order” symbol for X n1 .

Team logic

162

Let ξt1 =t2 be t1 = t2 and let ξ Rt1 ...tn be Rt1 . . . tn . For second order atomic formula, we let ξ X n1 t be t E x2n . Proceeding to compound formulas, let ξφ∨ψ = ξφ ∨ ξψ , ξφ∧ψ = ξφ ∧ ξψ , ξ¬φ = ∼ ξφ , ξ∃xn φ = ∃xn (P xn ∧ ξφ ), and finally ξ∃X n1 φ = ∃z n ξφ . Let θ be the conjunction of the (first order definable) sentence ∀x0 ∀x1 (¬x0 E x1 ∨ P x0 ) and the Axiom of Extensionality ∀x0 ∀x1 (∃x2 ((x2 E x0 ∧ ¬x2 E x1 ) ∨ (x2 E x1 ∧ ¬x2 E x0 )) ∨ x0 = x1 ). Let  be the following sentence of team logic: θ ∧ ∼ ∃x6 ! x0 ∃x1 ! x4 ∃x5 (= (x4 , x5 ) ∧ (¬x4 , x5 ) ∧ (¬x1 = x6 ⊗ P x0 ) ∧ (¬x0 = x5 ⊗ (x5 E x4 ∧ ¬x1 = x6 ) ⊗ (¬x5 E x4 ∧ x1 = x6 ))). The sentence  is true in a structure M if and only if M ∼ = N for some N such that E N = {(a, b) : a ∈ P N , a ∈ b ∈ N } and N = P(P N ). Now one can use induction on φ ∈ L 2 to prove that if N is as above then M |= φ if and only if N |= ξφ . More precisely, one proves for all formulas φ ∈ L 2 and all assignments s and S the following: M |=s,S φ if and only if N |= X s,S ξφ , where X s,S = {s }, s (x2n+1 ) = s(x2n+1 ), and s (x2n ) = S(X n1 ).



Corollary 8.17 A second order sentence φ has a model (of cardinality κ) if and only if the team logic sentence  ∧ ξφ has a model (of cardinality 2κ ). With the translation φ → ξφ we can consider second order logic L 2 as an implicitly defined fragment of team logic TL. Corollary 8.18 The Decision Problems of team logic and second order logic are recursively isomorphic.2 They have the same L¨owenheim and Hanf numbers. They have the same -extension. 2

Two sets of natural numbers are recursively isomorphic if there is a recursive bijection of N which maps one to the other.

8.6 Ehrenfeucht–Fra¨ıss´e game for team logic

163

The method of Theorem 8.16 was used in ref. [29] to prove that second order logic has an implicit translation into the extension of first order logic by the Henkin quantifier (see Exercise 6.20). As is apparent from the proof, we do not need the full power of team logic. It suffices to have the ∼-negated dependence logic sentence which forms one conjunct of . This emphasizes the strength of ∼ as compared to ¬. We note without proof that there is also a direct translation of second order logic into team logic, based on ref. [13]. Exercise 8.19 Suppose κ is a cardinal number with the following property: if φ ∈ TL, then every model M of TL has a submodel N such that N |= φ and |N | < κ. Show that κ is inaccessible. (In fact, κ is supercompact.)

8.6 Ehrenfeucht–Fra¨ıss´e game for team logic We now introduce an Ehrenfeucht–Fra¨ıss´e game adequate for team logic and use this game to characterize ≡TL . This game is simply a “two-directional” version of the Ehrenfeucht–Fra¨ıss´e game of D. Definition 8.19 Let M and N be two structures of the same vocabulary. The game EFTL n has two players and n moves. The position after move m is a pair (X, Y ), where X ⊆ M im and Y ⊆ N im for some i m . At the beginning, the position is ({∅}, {∅}) and i 0 = 0. Suppose the position after move number m is (X, Y ). There exist the following possibilities for the continuation of the game. Splitting move. Player I represents X (or Y ) as a union X = X 0 ∪ X 1 . Then player II represents Y (respectively, X ) as a union Y = Y0 ∪ Y1 . Now player I chooses whether the game continues from the position (X 0 , Y0 ) or from the position (X 1 , Y1 ). Duplication move. Player I decides that the game should continue from the new position, (X (M/xim ), Y (N /xim )). Supplementing move. Player I chooses a function F : X → M (or F : Y → N ). Then player II chooses a function G : Y → N (respectively, G : X → M). Then the game continues from the position (X (F/xim ), Y (G/xim )). After n moves, the position (X n , Yn ) is reached and the game ends. Player II is the winner if M |= X n φ ⇔ N |=Yn φ

164

Team logic

holds for all atomic, dual atomic, dependence, and dual dependence formulas φ(x0 , . . . , xin −1 ). Otherwise, player I wins. This is a game of perfect information and the concept of winning strategy is defined as usual. By the Gale–Stewart Theorem, the game is determined. Definition 8.20 (i) qrTL (φ) = 0 if φ is atomic, dual atomic, dependence or dual dependence formula. (ii) qrTL (φ ⊗ ψ) = max(qrTL (φ), qrTL (ψ))+1. (iii) qrTL (φ ∧ ψ) = max(qrTL (φ), qrTL (ψ)). (iv) qrTL (∃xn φ) = qrTL (φ)+1. (v) qrTL (!xn φ) = qrTL (φ)+1. (vi) qrTL (∼ φ) = qrTL (φ) Let FmlnTL,m be the set of formulas φ of dependence logic with qrTL φ ≤ m and with free variables among x0 , . . . , xn−1 . We write M ≡nTL N , if M |= φ is equivalent to N |= φ for all φ in Fml0TL,n . Note that there are for each n and m, up to logical equivalence, only finitely many formulas in FmlnTL,m . Theorem 8.21 Suppose M and N are models of the same vocabulary. Then the following are equivalent: (1) player II has a winning strategy in the game EFTL n (M, N ); (2) M ≡nTL N . Proof As in the proof of Theorem 6.44 we prove the equivalence, for all n, of the following two statements. (3)m Player II has a winning strategy in the game EFTL m (M, N ) in position n n (X, Y ), where X ⊆ M and Y ⊆ N . (4)m If φ is a formula in FmlnTL,m , then M |= X φ ⇔ N |=Y φ.

(8.1)

The proof is by induction on m. For each m we prove the claim simultaneously for all n. The case m = 0 is true by construction. Let us then assume (3)m ⇐⇒ (4)m as an induction hypothesis. Assume now (3)m+1 and let φ be a formula in FmlnTL,m+1 such that M |= X φ. As part of the induction hypothesis we assume that the claim in Eq. (8.1) holds for formulas shorter than φ. Case (1) φ = ψ0 ⊗ ψ1 , where ψ0 , ψ1 ∈ FmlnTL,m . Since M |= X φ, there are X 0 and X 1 such that X = X 0 ∪ X 1 , M |= X 0 ψ0 and M |= X 1 ψ1 . We let

8.6 Ehrenfeucht–Fra¨ıss´e game for team logic

Case (2)

Case (3) Case (4)

Case (5)

165

I play {X 0 , X 1 }. Then II plays according to her winning strategy, {Y0 , Y1 }. Since the next position in the game can be either one of (X 0 , Y0 ), (X 1 , Y1 ), we can apply the induction hypothesis to both. This yields N |=Y0 ψ0 and N |=Y1 ψ1 . Thus N |=Y φ. TL,m φ = ∃xn ψ, where ψ ∈ Fmln−1 . Since M |= X φ, there is a function F : X → M such that M |= X (F/xn ) ψ. We let I play F. Then II plays according to her winning strategy a function G : Y → N and the game continues in position (X (F/xn ), Y (G/xn )). The induction hypothesis gives N |=Y (G/xn ) ψ. Now N |=Y φ follows. φ = ∼ ψ. Since M |= X φ, we have M |= X ψ. By the induction hypothesis, N |=Y ψ. Thus N |=Y φ. φ = ψ0 ∧ ψ1 , where ψ0 , ψ1 ∈ FmlnTL,m . Since M |= X φ, we have M |= X ψ0 and M |= X ψ1 . By the induction hypothesis, N |=Y ψ0 and N |=Y ψ1 . Thus N |=Y φ. TL,m φ = !xn ψ, where ψ ∈ Fmln+1 . By assumption, M |= X (M/xn ) ψ. We let now I demand that the game continues in the new position (X (M/xn ), Y (N /xn )). The induction hypothesis gives N |=Y (N /xn ) ψ. Now N |=Y φ follows trivially.

To prove the converse implication, assume (4)m+1 . To prove (3)m+1 , we consider the possible moves that player I can make in the position (X, Y ). Case (i) Player I writes X = X 0 ∪ X 1 . Let φ j , j < k, be a complete list (up to logical equivalence) of formulas in FmlnTL,m . Since  M |= X 0 φj M|= X 0 φ j

and M |= X 1



φj,

M|= X 1 φ j

we have

⎛

⎞



M |= X ⎝

⎛



φj⎠ ∨ ⎝

M|= X 0 φ j

⎞ φj⎠ .

M|= X 1 φ j

Note that  TL

qr

 M|= X 1 φ j

 φj

= max (φ j ) ≤ m. M|= X 1 φ j

Team logic

166

Therefore by (4)m+1 ⎛  N |=Y ⎝

⎞

⎛

φj⎠ ∨ ⎝

M|= X 0 φ j

⎞



φj⎠ .

M|= X 1 φ j

Thus Y = Y0 ∪ Y1 such that N |=Y0



φj

M|= X 0 φ j

and 

N |=Y1

φj.

M|= X 1 φ j

By this and the induction hypothesis, player II has a winning strategy in the positions (X 0 , Y0 ), (X 1 , Y1 ). Thus she can play {Y0 , Y1 } and maintain her winning strategy. Case (ii) Player I decides that the game should continue from the new position (X (M/xn ), Y (m/xn )). We claim that M |= X (M/xn ) φ ⇔ N |=Y (N /xn ) φ TL,m for all φ ∈ Fmln+1 . From this the induction hypothesis would imply that II has a winning strategy in the position (X (M/xn ), Y (N /xn )). TL,m So let us assume M |= X (M/xn ) φ, where φ ∈ Fmln+1 . By definition,

M |= X ! xn φ. ,m+1 Since ! xn φ ∈ FmlTL , (4)m+1 gives N |=Y ! xn φ and N |=Y (N /xn ) n φ follows. Case (iii) Player I chooses a function F : X → M. Let φi , i < M be a TL,m complete list (up to logical equivalence) of formulas in Fmln+1 . Now  M |= X ∃xn φi . M|= X (F/xn ) φi

Note that  TL qr ∃xn



 φj

 = qr

M|= X (F/xn ) φ j

 =



TL

M|= X (F/xn ) φ j

max

M|= X (F/xn ) φ j

 φ j +1

 qrTL (φ j ) +1 ≤ m+1

8.6 Ehrenfeucht–Fra¨ıss´e game for team logic

and hence, by (4)m+1 , N |=Y ∃xn



167

φi .

M|= X (F/xn ) φi

Thus there is a function G : Y → N such that  N |=Y (G/xn ) φi . M|= X (F/xn ) φi

The game continues from position (X (F/xn ), Y (G/xn )). Given that now M |= X (F/xn ) φ ⇔ N |=Y (G/xn ) φ TL,m for all φ ∈ Fmln+1 , the induction hypothesis implies that II has a winning strategy in position (X (F/xn ), Y (G/xn )). 

Corollary 8.22 Suppose M and N are models of the same vocabulary. Then the following are equivalent: (1) M ≡TL N ; (2) for all natural numbers n, player II has a winning strategy in the game EFTL n (M, N ). Proposition 8.23 Suppose K is a model class and n is a natural number. Then the following are equivalent: (1) K is definable in team logic by a sentence in Fml0TL,n ; (2) K is closed under the relation ≡nTL . Proof Suppose K is the class of models of φ ∈ Fml0TL,n . If M |= φ and M ≡nTL N , then, by definition, N |= φ. Conversely, suppose K is closed under ≡nTL . Let  φM = {φ : φ ∈ Fml0TL,n , M |= φ}, where the conjunction is taken over a finite set which covers all such φ up to logical equivalence. Let θ be the disjunction of all φM , where M ∈ K . Again we take the disjunction over a finite set up to logical equivalence. We show that K is the class of models of θ. If M ∈ K , then M |= φM , whence M |= θ . On the other hand, suppose M |= φN for some N ∈ K . Now N ≡nTL M, for if N |= φ and φ ∈ Fml∅TL,n , then φ is logically equivalent with one of the conjuncts of φN , whence M |= φ. As K is closed under ≡nTL , we have M ∈ K . 

168

Team logic

Corollary 8.24 Suppose K is a model class. Then the following are equivalent: (1) K is definable in team logic; (2) there is a natural number n such that K is closed under the relation MRN ⇐⇒ player II has a winning strategy in EFTL n (M, N ). We have obtained, after all, a purely game theoretic definition of team logic. Exercise 8.20 Describe an Ehrenfeucht–Fra¨ıss´e game for the fragment of team logic in which ⊗ does not occur. Give a game theoretic characterization in the spirit of Corollary 8.24 of this fragment.

Appendix Solutions to selected exercises, by Ville Nurmi

Chapter 2 Exercise 2.1 Let M be a model and let M be its universe. The task is to prove that for each first order φ and each assignment s with Fr(φ) ⊆ dom(s) it holds that either (φ, s, 1) ∈ T or (φ, s, 0) ∈ T . We prove it by induction on φ. Assume φ is of form t1 = t2 . It is a fact that either t1M s = t2M s or t1M s = t2M s. Thus, by definition of T , either (t1 = t2 , s, 1) ∈ T or (t1 = t2 , s, 0) ∈ T . Assume φ is of form Rt1 . . . tn . Again, either (t1M s, . . . , tnM s) ∈ R M or M (t1 s, . . . , tnM s) ∈ R M . Therefore, by the definition of T , either (Rt1 . . . tn , s, 1) ∈ T or (Rt1 . . . tn , s, 0) ∈ T . For the following inductive steps we assume as our induction hypothesis that the claim holds for subformulas of φ. Assume φ is of form ¬ψ. By the induction hypothesis, either (ψ, s, 1) ∈ T or (ψ, s, 0) ∈ T . By definition of T , we get in the first case that (¬ψ, s, 0) ∈ T , and in the latter case (¬ψ, s, 1) ∈ T . Assume φ is of form ψ ∨ θ . If (ψ, s, 0) ∈ T and (θ, s, 0) ∈ T , then by definition of T also (ψ ∨ θ, s, 0) ∈ T . Otherwise, either (ψ, s, 0) ∈ T or (θ, s, 0) ∈ T , whence, by the induction hypothesis, either (ψ, s, 1) ∈ T or (θ, s, 1) ∈ T . By definition of T , we get (ψ ∨ θ, s, 1) ∈ T . Finally, assume φ is of form ∃xn ψ. If there is some a ∈ M satisfying (ψ, s(a/xn ), 1) ∈ T , then, by definition of T , we get (∃xn ψ, s, 1) ∈ T . Otherwise, for all a ∈ M we have (ψ, s(a/xn ), 1) ∈ T . By the induction hypothesis, for all a ∈ M we have (ψ, s(a/xn ), 0) ∈ T , whence, by definition of T , (∃xn ψ, s, 0) ∈ T .

Exercise 2.7 The task is to show φ ≡ φ and ¬φ ≡ φ d for all first order φ. We prove both claims simultaneously by induction on φ. The induction hypothesis is that the claim holds for subformulas of φ. In the following, the places where the induction hypothesis is used are marked by IH. p

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170

If φ is atomic, then φ p = φ and φ d = ¬φ, by definition. IH

IH

If φ is of form ¬ψ, then (¬ψ)p = ψ d ≡ ¬ψ. Also (¬ψ)d = ψ p ≡ ψ ≡ ¬¬ψ. IH

If φ is of form ψ ∨ θ, then (ψ ∨ θ)p = ψ p ∨ θ p ≡ ψ ∨ θ. Also, (ψ ∨ θ)d = ψ d ∧ d IH

θ ≡ (¬ψ) ∧ (¬θ ) ≡ ¬(ψ ∨ θ ). IH

Finally, if φ is of form ∃xn ψ, then (∃xn ψ)p = ∃xn ψ p ≡ ∃xn ψ. Also, (∃xn ψ)d = d IH

∀xn ψ ≡ ∀xn (¬ψ) ≡ ¬∃xn ψ.

Chapter 3 Exercise 3.2 Let X ∈ Team(M, {x0 , x1 }), i.e., let X be some team of assignments s : {0, 1} → M. For X to be of type φ in M it means that M |= X φ. Now M |= X =(x0 , x1 ) ⇐⇒ (=(x0 , x1 ), X, 1) ∈ T . By Proposition 3.8, this is equivalent to the condition that for all s, s  ∈ X it holds that if s(x0 ) = s  (x0 ) then s(x1 ) = s  (x1 ). An alternative characterization can be given in terms of a function f : A → M for some A ⊆  M. Then a team X of type =(x0 , x1 ) is of the form {(0, a), (1, f (a))} : a ∈ A for some A and f . Similarly to the previous case, M |= X =(x1 , x0 ) ⇐⇒ (=(x1 , x0 ), X, 1) ∈ T ⇐⇒  (by Proposition 3.8) for all s, s  ∈ X , it holds that if s(x1 ) = s  (x1 ) then s(x  0 ) = s (x0 ). In other words, such teams X are of the form {(0, f (a)), (1, a)} : a ∈ A for some set A ⊆ M and function f : M → M. M |= X =(x0 , x0 ) ⇐⇒ (=(x0 , x0 ), X, 1) ∈ T ⇐⇒ (by Proposition 3.8) for all s, s  ∈ X it holds that if s(x0 ) = s  (x0 ) then s(x0 ) = s  (x0 ). But this is a trivial condition, satisfied by any team X . Thus all X ∈ Team(M, {x0 , x1 }) are of type =(x0 , x0 ).

Exercise 3.3 The language consists of a constant symbol c and a function symbol f . We consider teams X ∈ Team(M, {x0 }). M |= X =(c, c) ⇐⇒ (by Proposition 3.8) for all s, s  ∈ X it holds that if cM = cM then cM = cM . This condition is trivially true for all X . Thus all teams X ∈ Team(M, {x0 }) are of type =(c, c). M |= X =(x0 , c) ⇐⇒ (by Proposition 3.8) for all s, s  ∈ X , it holds that if s(x0 ) = s  (x0 ) then cM = cM . This condition is also trivially true, so all teams X ∈ Team(M, {x0 }) are of type =(x0 , c). M |= X =(c, x0 ) ⇐⇒ (by Proposition 3.8) for all s, s  ∈ X , it holds that if cM = cM then s(x0 ) = s  (x0 ). As cM = cM always holds, the condition states that there is some fixed a ∈ M such that all s ∈ X map s(x0 ) = a. Considering also that dom(s) = {0} for  all s ∈ X , we get that either X = ∅ or it is of the form {(0, a)} for some a ∈ M, i.e., there is at most one agent in team X . M |= X =(c, f x0 ) ⇐⇒ (by Proposition 3.8) for all s, s  ∈ X , it holds that if cM = M c then f M (s(x0 )) = f M (s  (x0 )) ⇐⇒ for all s, s  ∈ X : f M (s(x0 )) = f M (s  (x0 )) ⇐⇒ f M is constant in the set {s(x0 ) : s ∈ X }.

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Table A.1. a

b

a+b

a·b

a

b

a+b

a·b

2 2 2 2 3

2 3 4 5 3

4 5 6 7 9

4 6 8 10 9

3 3 4 4 5

4 5 4 5 5

7 8 8 9 10

12 15 16 20 25

Exercise 3.5 Here the language has two function symbols, f and g, and we consider teams X ∈ Team(M, {x0 , x1 }). M |= X =( f x0 , x0 ) ⇐⇒ (by Proposition 3.8) for all s, s  ∈ X , it holds that if M f (s(x0 )) = f M (s  (x0 )) then s(x0 ) = s  (x0 ) ⇐⇒ the function f M {s(x0 ) : s ∈ X } is one-to-one. One can also characterize this condition by saying, a bit vaguely, that f M is one-to-one in {s(x0 ) : s ∈ X }. M |= X =( f x1 , x0 ) ⇐⇒ (by Proposition 3.8) for all s, s  ∈ X , it holds that if M f (s(x1 )) = f M (s  (x1 )) then s(x0) = s  (x0 ). This is equivalent to being able to define a function h : f M (s(x1 )) : s ∈ X → M by the equation h( f M (s(x1 ))) = s(x0 ). M |= X =( f x0 , gx1 ) ⇐⇒ (by Proposition 3.8) for all s, s  ∈ X , it holds that if f M (s(x0 )) = f M (s  (x0 )) then g M (s(x1 )) = g M (s (x1 )). This is equivalent to being able to define a function h : f M (s(x0 )) : s ∈ X by the equation h( f M (s(x0 ))) = g M (s(x1 )).

Exercise 3.7

  Let X 5 = {(0, a), (1, b)}: 1 < a ≤ n, 1 < b ≤ n, a ≤ b . It is possible to prove in general that, for any natural number n, X n is of type =(x0 + x1 , x0 · x1 , x0 ). When inspecting just the case n = 5 we can save ourselves the trouble of the general proof and  just check all cases as they are not that many. Thus we list for each element s = {(0, a), (1, b)} ∈ X 5 the values a, b, a + b, and a · b (see Table A.1). We can see that for no two s, s  ∈ X 5 does it hold that s(x0 ) + s(x1 ) = s  (x0 ) +  s (x1 ) and s(x0 ) · s(x1 ) = s  (x0 ) · s  (x1 ) and s(x0 ) = s  (x0 ). This proves that X 5 is of type =(x0 + x1 , x0 · x1 , x0 ). Here is an alternative, general proof that X n is of type =(x0 + x1 , x0 · x1 , x0 ), for any n. Let 1 < x0 ≤ x1 ≤ n. Denote a = x0 + x1 and b = x0 · x1 . Then we have a 2 − 4b = x02 − 2x0 x1 + x12 = (x1 − x0 )2 ≥ 0. To see that knowing the values of a and b gives us a unique value for x0 , we can continue as follows. Because 2 x1 = a − x0 , we get b = −x02 + ax √ 0 . This gives the quadratic equation x0 − ax0 + b = 0, which yields x0 = 12 (a ± a 2 − 4b). Because x0 ≤ x1 , the only solution is √ x0 = 12 (a − a 2 − 4b).

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Exercise 3.10 In order to show |= φ for some sentence φ, one must show that, for all models M, M |={∅} φ. Therefore, let M be arbitrary: M |={∅} ∀x0 ∀x1 (x1 = f x0 → =(x0 , x1 )) ⇔ M |= X ¬x1 = f x0 ∨ =(x0 , x1 ), where X = {s : {0, 1} → M} ⇔ there are teams Y, Z such that X = Y ∪ Z , dom(Y ) = dom(Z ), and M |=Y ¬x1 = f x0 and M |= Z =(x0 , x1 ).

(A.1)

So, in order to prove the claim, it suffices to find teams Y and Z satisfying the condition given in Eq. (A.1). We can choose Y = {s ∈ X : s(x1 ) = f M (s(x0 ))} and Z = X \ Y . Then, clearly, X = Y ∪ Z and dom(Y ) = dom(Z ). It is easy to check that M |=Y ¬x1 = f (x0 ). Also M |= Z =(x0 , x1 ) because, in fact, Z = {s ∈ X : s(x1 ) = f M (s(x0 ))}, so, whenever we have some s, s  ∈ Z with s(x0 ) = s  (x0 ), then s(x1 ) = f M (s(x0 )) = f M (s  (x0 )) = s  (x1 ).

Exercise 3.11 First note that for all formulas φ and models M it holds that (φ, ∅, 1) ∈ T and (¬φ, ∅, 1) ∈ T , i.e., M |=∅ φ and M |=∅ ¬φ. This means that M |= X ¬φ ⇒ M |= X φ fails for X = ∅. Therefore, when studying the claim M |= X ¬φ ⇔ M |= X φ for some model M and team X , we limit ourselves to non-empty teams. Let φ be the formula =(x0 , x1 ) ∧ ¬x0 = x1 . Assume towards contradiction that for all models M and teams X = ∅ it holds that M |= X ¬φ ⇐⇒ M |= X φ. Let us inspect the model M of the empty language, with universe M = {a, b}, where a and b are two different elements. (There are also lots of other possible models to consider, but this is probably the simplest one.) Let X = {s, s  }, where s = {(0, a), (1, a)} and s  = {(0, a), (1, b)}. Then M |= X φ because M |= X =(x0 , x1 ). By assumption, M |= X ¬φ. By the Closure Test, M |={s  } ¬φ. Again, by assumption, M |={s  } φ. But this is a contradiction with the facts that M |={s  } =(x0 , x1 ) and M |={s  } ¬x0 = x1 . Thus the equivalence does not hold for =(x0 , x1 ) ∧ ¬x0 = x1 . Let ψ be the formula =(x0 , x1 ) → x0 = x1 , i.e. ¬ =(x0 , x1 ) ∨ x0 = x1 . Assume towards contradiction that for all models M and teams X = ∅ it holds that M |= X ¬ψ ⇐⇒ M |= X ψ. Let M and X = {s, s  } be as earlier. Now M |= X ψ because for all X = Y ∪ Z either M |=Y ¬ =(x0 , x1 ) or M |= Z x0 = x1 . Namely, M |= Z x0 = x1 can hold only for Z = ∅ and for Z = {s} (assuming that Z ⊆ X ), and, by Proposition 3.8, M |=Y ¬ =(x0 , x1 ) can hold only when Y = ∅. We cannot have X = Y ∪ Z with any sets Y and Z satisfying these conditions. Now, by the assumption, M |= X ¬ψ. By the Closure Test, M |={s} ¬ψ. Again by the assumption, M |={s} ψ. But this cannot be true because we can split {s} = ∅ ∪ {s}, and it holds that M |=∅ ¬ =(x0 , x1 ) and M |={s} x0 = x1 . We have ended up in a contradiction, so M |= X ψ ⇔ M |= X ¬ψ does not hold for all models M and teams X = ∅. Let θ be the formula =(x0 , x1 ) ∨ ¬x0 = x1 . By definition, M |= X ¬θ ⇐⇒ (¬θ, X, 1) ∈ T ⇐⇒ (θ, X, 0) ∈ T ⇐⇒ (=(x0 , x1 ), X, 0) ∈ T and (¬x0 = x1 , X, 0) ∈ T ⇐⇒ X = ∅ (by Proposition 3.8, (=(x0 , x1 ), X, 0) ∈ T iff X = ∅), and we have excluded this case from our investigations. Therefore M |= X ¬θ fails for all models M and teams X = ∅. On the other hand, for any model M and team

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173

X = ∅, letting Z = {s ∈ X : s(x0 ) = s(x1 )} and Y = X \ Z , we see that X = Y ∪ Z , dom(Y ) = dom(Z ), M |=Y =(x0 , x1 ), and M |= Z ¬x0 = x1 . Therefore M |= X θ fails for all M and X = ∅. Hereby we have proven the equivalence M |= X θ ⇔ M |= X ¬θ for all models M and teams X = ∅.

Exercise 3.13 Let M be a model. Assume that each Ti for i ∈ Isatisfies conditions (D1)–(D12) of Definition 3.5. Our goal is to show that in that case i∈I Ti satisfies the conditions also. The proof proceeds condition by condition. At each step we consider some arbitrary X ∈ Team(M, V ), where V contains all variable indices that appear in the formula in question. (D1) Assume that for all s ∈ X we have t1M s = t2M s. Then, because each  Ti satisfies (D1), we have (t = t , X, 1) ∈ T . Therefore (t = t , X, 1) ∈ 1 2 i 1 2 i∈I Ti , so  i∈I Ti satisfies (D1). (D2) Assume that for all s ∈ X we have t1M s = t2M s. Then, because each  Ti satisfies (D2), we have (t1 = t2 , X, 0) ∈ Ti . Therefore (t1 = t2 , X, 0) ∈ i∈I Ti , so  i∈I Ti satisfies (D2). (D3) Assume that for all s, s  ∈ X it holds that if tiM s = tiM s   for i < n, then tnM s = tnM s  . Then, because each Ti satisfies  (D3), we  have (=(t1 , . . . , tn ), X, 1) ∈ Ti . Therefore (=(t1 , . . . , tn ), X, 1) ∈ i∈I Ti , so i∈I Ti satisfies (D3). (D4) Because each Ti satisfies  (D4), we  have (=(t1 , . . . , tn ), ∅, 0) ∈ Ti . Therefore (=(t1 , . . . , tn ), ∅, 0) ∈ i∈I Ti , so i∈I Ti satisfies (D4). (D5) Assume that for all s ∈ X we have (t1M s, . . . , tnM s) ∈ R M . Then, because each Ti satisfies (D5),  we have(Rt1 . . . tn , X, 1) ∈ Ti . Therefore the triple (Rt1 . . . tn , X, 1) is in i∈I Ti , so i∈I Ti satisfies (D5). (D6) Assume that for all s ∈ X we have (t1M s, . . . , tnM s) ∈ R M . Then, because each Ti satisfies (D6),  we have(Rt1 . . . tn , X, 0) ∈ Ti . Therefore the triple (Rt1 . . . tn , X, 0) is in i∈I Ti , so i∈I Ti satisfies (D6).  (D7) Assume that (φ, X, 1) ∈ i∈I Ti and (ψ, Y, 1) ∈ i∈I Ti and dom(X ) = dom(Y ). Then (φ, X, 1) ∈ Ti and (ψ, Y, 1) ∈ Ti for each i ∈ I . Because  Ti satisfies (D7), (φ ∨ ψ, X ∪ Y, 1) ∈ T . Therefore (φ ∨ ψ, X ∪ Y, 1) ∈ i i∈I Ti , so  T satisfies (D7). i i∈I   (D8) Assume that (φ, X, 0) ∈ i∈I Ti and (ψ, X, 0) ∈ i∈I Ti . Then (φ, X, 0) ∈ Ti and (ψ, X, 0) ∈ Ti for each i ∈ I . As  Ti satisfies  (D8), (φ ∨ ψ, X, 0) ∈ Ti for each i ∈ I . Therefore (φ ∨ ψ, X, 0) ∈ T , so i i∈I i∈I Ti satisfies (D8).  (D9) Assume that (φ, X, 1) ∈ i∈I Ti . Then (φ, X, 1) ∈ Ti for each i ∈  I . Because each T satisfies (D9), (¬φ, X, 0) ∈ T . Therefore (¬φ, X, 0) ∈ i i i∈I Ti , so  i∈I Ti satisfies (D9).  (D10) Assume that (φ, X, 0) ∈ i∈I Ti . Then (φ, X, 0) ∈ Ti for each i ∈  I . Because each T satisfies (D10), (¬φ, X, 1) ∈ T . Therefore (¬φ, X, 1) ∈ i i i∈I Ti , so  i∈I Ti satisfies (D10).  (D11) Let F : X → M, and assume that (φ, X (F/xn ), 1) ∈ i∈I Ti . Then (φ, X (F/xn ), 1) ∈ Ti for each i ∈ I . Because satisfies (D11),  each Ti  (∃xn φ, X, 1) ∈ Ti . Therefore (∃xn φ, X, 1) ∈ i∈I Ti , so i∈I Ti satisfies (D11).

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Table A.2. The teams X and X (F/x0 ) for Exercise 3.15 X x0 s s

X (F/x0 ) x1

x2

a b

a a

s s

x0

x1

x2

c c

a b

a a

 (D12) Assume that (φ, X (M/xn ), 0) ∈ i∈I Ti . Then (φ, X (M/xn ), 0) ∈ Ti for each i ∈ I . Because each   Ti satisfies (D12), (∃xn φ, X, 0) ∈ Ti . Therefore (∃xn φ, X, 0) ∈ i∈I Ti , so i∈I Ti satisfies (D12).

Exercise 3.15



 Denote by ψ the formula ∃x0 =(x2 , x0 ) ∧ ¬(x0 = x1 ) and by φ the formula   ∃x0 =(x2 , x0 ) ∧ x0 = x1 . To show that M |= X φ ⇐⇒ M |= X ψ does not hold, let M = {a, b, c}, where a, b, and c are three different elements, and let X = {s, s  } ∈ Team(M, {x1 , x2 }), where s = {(x1 , a), (x2 , a)} and s  = {(x1 , b), (x2 , a)} (see Table A.2). Then M |= X ψ and M |= X φ. We can see this by noting that x2 is constant in X . Because, on the other hand, both φ and ψ contain =(x2 , x0 ), any function F : X → M that unravels ∃x0 in φ or ψ must be a constant function. We can choose a constant that avoids both a and b, but we cannot choose a constant that equals both a and b. More precisely, let F : X → M map F(s) = c for all s ∈ X . Then M |= X (F/x0 ) ¬(x0 = x1 ), and M |= X (F/x0 ) =(x2 , x0 ). Thus M |= X (F/x0 ) ¬(x0 = x1 ) ∧ =(x2 , x0 ), so M |= X ψ. On the other hand, if M |= X φ was to hold, there would be some G : X → M such that M |= X (G/x0 ) =(x2 , x0 ) and M |= X (G/x0 ) x0 = x1 . From the former we get that when s, s  ∈ X (G/x0 ) and s(x2 ) = s  (x2 ), then s(x0 ) = s  (x0 ). But s(x2 ) = s  (x2 ) holds for all s, s  ∈ X (G/x0 ). Therefore G must be a constant function; there is some x ∈ M such that for all s ∈ X : G(s) = x. But then by M |= X (G/x0 ) x0 = x1 it must be that x = a, and x = b, which is a contradiction. Thus M |= X φ. 

Exercise 3.17

 Denote by φ the formula ∃x1 =( f x0 , x1 ) ∧ f x1 = f x0 . We want to falsify the following claim: M |= X φ ⇐⇒ f M is one-to-one in {s(x0 ) : s ∈ X }. Let M = {a, b}, where a and b are distinct elements, interpret f M (x) = a for all x ∈ M, and let X = {s, s  } ∈ Team(M, {x0 }), where s = {(0, a)} and s  = {(0, b)} (see Table A.3). Clearly, f M is not one-to-one in {a, b} because it is a constant function, but M |= X φ; namely, let F : X → M map F(s) = a for all s ∈ X . Then for all s ∈ X (F/x1 ) it holds that f M (s(x1 )) = f M (s(x0 )) simply because f M is a constant function, and thus M |= X (F/x1 ) f x1 = f x0 . Also, when s, s  ∈ X (F/x1 ), it holds that s(x1 ) = s  (x1 ), so M |= X (F/x1 ) =( f x0 , x1 ). Then M |= X (F/x1 ) =( f x0 , x1 ) ∧ f x1 = f x0 , so M |= X φ.

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Table A.3. The teams X and X (F/x1 ) for Exercise 3.17 X x0 s s

X (F/x1 ) x1

a b

s s

x0

x1

a b

a a

Table A.4. The teams X and X (F/x1 ) for Exercise 3.18 X x0 s

a

X (F/x1 ) x1 s

x0

x1

a

a

In fact, it is possible to prove that any team is of type φ in any model. Then by simply defining a model where f M is not one-to-one we can falsify the claim. 

Exercise 3.18

 Denote by φ the formula ∃x1 =(x0 , x1 ) ∧ Rx0 x1 . We want to falsify the following claim: M |= X φ ⇐⇒ R M is a function and dom(R M ) = {s(x0 ) : s ∈ X }. Let M = {a, b} have two distinct elements, let R M = {(a, a), (a, b)}, and let X = {s} ∈ Team(M, {x0 }), where s = {(0, a)} (see Table A.4). Clearly, R M is not a function with domain {a}, but M |= X φ; namely, let F : X → M map F(s) = a. Then M |= X (F/x1 ) =(x0 , x1 ) simply because X (F/x1 ) is a singleton. Also M |= X (F/x1 ) Rx0 x1 because X (F/x1 ) = {{(0, a), (1, a)}} and (a, a) ∈ R M . Thus M |= X (F/x1 ) =(x0 , x1 ) ∧ Rx0 x1 , so M |= X φ.

Exercise 3.21 We prove only items (i) and (ii). (i) Our first goal is to find some D-formula φ such that φ ∨ ¬φ ≡∗  fails. Let M = {a, b} be a two-element universe and X = {s, s  } ∈ Team(M, {x0 }), where s : 0 → a, and s  : 0 → b. Let us investigate the formula =(x0 ). If a team is of type =(x0 ), then x0 is constant over the team. Now, by Proposition 3.8, for a team Y ∈ Team(M, {x0 }) it holds that M |=Y =(x0 ) iff for all s, s  ∈ Y : s(x0 ) = s  (x0 ) iff Y has at most one agent. Also, for a team Z ∈

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Team(M, {x0 }) it holds that M |= Z ¬ =(x0 ) iff Z = ∅. Therefore we cannot split X into some Y ∪ Z so that M |=Y =(x0 ) and M |= Z ¬ =(x0 ). Thus M |= X =(x0 ) ∨ ¬ =(x0 ). However, it is the case that M |= X . Hereby we see that the equivalence M |= X φ ∨ ¬φ ⇐⇒ M |= X  fails, when we choose φ to be =(x0 ). Thus φ ∨ ¬φ ≡  fails, and furthermore φ ∨ ¬φ ≡∗  fails. We have, by definition, φ ∧ ¬φ = ¬(φ ∨ ¬φ) and ⊥ = ¬. Therefore the failure of φ ∨ ¬φ ≡∗  gives us the failure of φ ∧ ¬φ ≡∗ ⊥. Let us then prove φ ∧ ¬φ ≡ ⊥. It is shown simply by letting M be an arbitrary model, letting X be an arbitrary team, and letting φ be an arbitrary D-formula. Then we apply Proposition 3.8: (φ ∧ ¬φ, X, 1) ∈ T ⇐⇒ (φ, X, 1) ∈ T and (¬φ, X, 1) ∈ T ⇐⇒ (φ, X, 1) ∈ T and (φ, X, 0) ∈ T ⇐⇒ (⊥, X, 1) ∈ T . The final equality holds because both sides are true only when X = ∅. [Note: When we speak about teams X ∈ Team(M, {x0 }), we can use a shorthand notation and write X ⊆ M. That is, we can think that a team that interprets only one variable is a subset of the universe. In reality, this is never the case, but the shorthand notation X ⊆ M would instead mean the team X  = {sa : a ∈ X }, where sa = {(x0 , a)} for each a ∈ X . The shorthand notation is okay if we are certain which variable the team is meant to interpret. When we talk about a formula with one free variable, we can be certain of the variable.] (ii) Let us then prove φ ∧ φ ≡ φ. Let M be an arbitrary model, let X be an arbitrary team, and let φ be an arbitrary D-formula. We apply Proposition 3.8: (φ ∧ φ, X, 1) ∈ T ⇐⇒ (φ, X, 1) ∈ T and (φ, X, 1) ∈ T ⇐⇒ (φ, X, 1) ∈ T . To show that φ ∧ φ ≡∗ φ it suffices to find some model M, team X , and D-formula φ for which (φ ∧ φ, X, 0) ∈ T and (φ, X, 0) ∈ T . To this end, let M = {a, b} be a twoelement universe, X = {s, s  } ∈ Team(M, {x0 }), where s = {(0, a)} and s  = {(0, b)} (or with shorthand notation, X = {a, b}), and let φ be the formula ¬ =(x0 ). Now, because there are s, s  ∈ X with s(x0 ) = s  (x0 ), by Proposition 3.8 (=(x0 ), X, 1) ∈ T , and so (φ, X, 0) ∈ T . But, on the other hand, any team containing at most one agent is of the type of any determination formula = (t1 , . . . , tn ). This can be seen from Proposition 3.8. In particular, (=(x0 ), {s}, 1) ∈ T , so (φ, {s}, 0) ∈ T , and likewise also (φ, {s  }, 0) ∈ T . Thus, by the truth definition, (φ ∧ φ, {s} ∪ {s  }, 0) ∈ T , so we get (φ ∧ φ, X, 0) ∈ T , which completes the proof.

Exercise 3.23 We prove only item (iv). First we will show (φ ∧ ψ) ∧ θ ≡∗ φ ∧ (ψ ∧ θ). Let φ, ψ, and θ be D-formulas, let M be a model, and let X ∈ Team(M, V ), where V = Fr(φ) ∪

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Fr(ψ) ∪ Fr(θ ). Then   (φ ∧ ψ) ∧ θ, X, 1 ∈ T ⇐⇒ (φ ∧ ψ, X, 1) ∈ T and (θ, X, 1) ∈ T ⇐⇒ (φ, X, 1) ∈ T and (ψ, X, 1) ∈ T and (θ, X, 1) ∈ T ⇐⇒ (φ, X, 1) ∈ T and (ψ ∧ θ, X, 1) ∈ T   ⇐⇒ φ ∧ (ψ ∧ θ ), X, 1 ∈ T . Also, 

 (φ ∧ ψ) ∧ θ, X, 0 ∈ T

⇐⇒ (φ ∧ ψ, Y, 0) ∈ T and (θ, Z , 0) ∈ T for some Y ∪ Z = X ⇐⇒ (φ, A, 0) ∈ T and (ψ, B, 0) ∈ T and (θ, Z , 0) ∈ T for some A ∪ B = Y ⇐⇒ (φ, A, 0) ∈ T and (ψ ∧ θ, B ∪ Z , 0) ∈ T   ⇐⇒ φ ∧ (ψ ∧ θ), A ∪ B ∪ Z , 0 ∈ T . Note that A ∪ B ∪ Z = X , so this completes the proof. Then we will show (φ ∨ ψ) ∨ θ ≡∗ φ ∨ (ψ ∨ θ). We can prove this with a proof similar to the one above. We can also do it using the Preservation of Equivalence Under Substitution. The identities below are by definition of ∨ in terms of ¬ and ∧. The strong equivalences, ≡∗ , used in the following proof are based on Lemma 3.22, which we proved above:   (φ ∨ ψ) ∨ θ = ¬(¬φ ∧ ¬ψ) ∨ θ = ¬ ¬¬(¬φ ∧ ¬ψ) ∧ ¬θ     ≡∗ ¬ (¬φ ∧ ¬ψ) ∧ ¬θ ≡∗ ¬ ¬φ ∧ (¬ψ ∧ ¬θ)   ≡∗ ¬ ¬φ ∧ ¬¬(¬ψ ∧ ¬θ ) = φ ∨ ¬(¬ψ ∧ ¬θ ) = φ ∨ (ψ ∨ θ).

Exercise 3.24 We prove only item (iv), starting from (iv)(a). Let φ be a D-formula. We have to show φ ⇒∗ ∃xn φ. In the following solution, it is essential to distinguish between two cases, whether xn appears free in φ or not. As we remember from the definition of strong consequence, we have to consider all teams X with Fr(φ) ∪ Fr(∃xn φ) ⊆ dom(X ). Because Fr(∃xn φ) ⊆ Fr(φ), the condition reduces to Fr(φ) ⊆ dom(X ). The distinction if xn appears free in φ shows up in the possible domains of the teams. Let M be a model. Assume that xn appears free in φ. We first show φ ⇒ ∃xn φ, so let X be a team with Fr(φ) ⊆ dom(X ), and assume (φ, X, 1) ∈ T . Because we assumed that n ∈ dom(X ), we can write X = X (F/xn ), where F : X → M is defined by F(s) = s(xn ). Thus (φ, X (F/xn ), 1) ∈ T , so (∃xn φ, X, 1) ∈ T . Then we show ¬∃xn φ ⇒ ¬φ, so let X have Fr(φ) ⊆ dom(X ), and assume (∃xn φ, X, 0) ∈ T . Then (φ, X (M/xn ), 0) ∈ T , or, in other words, (¬φ, X (M/xn ), 1) ∈ T . Because we assumed that n ∈ dom(X ), we have X ⊆ X (M/xn ), namely we can write any s ∈ X as s = s(s(xn )/xn ). Then, by the Closure Test we get (¬φ, X, 1) ∈ T , so (φ, X, 0) ∈ T . This completes the proof that φ ⇒∗ ∃xn φ in the case that xn is free in φ.

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Assume now that xn does not appear free in φ. In addition to the proof that we presented for the previous case, we also have to take into account teams X with xn ∈ dom(X ). With these teams the preceding proof does not work because we cannot write things like s = s(s(xn )/xn ) for all s ∈ X . We need different tricks. First we show φ ⇒ ∃xn φ, so let X have Fr(φ) ⊆ dom(X ), n ∈ dom(X ), and assume (φ, X, 1) ∈ T . Let a ∈ M, and define F : X → M by F(s) = a for all s ∈ X . Then we have (φ, X (F/xn ), 1) ∈ T because of Lemma 3.27 and we can write X = X (F/xn ) dom(X ). Now we get (∃xn φ, X, 1) ∈T. Then we show ¬∃xn φ ⇒ ¬φ, so let X have Fr(φ) ⊆ dom(X ), n ∈ dom(X ), and assume (∃xn φ, X, 0) ∈ T . Then (φ, X (M/xn ), 0) ∈ T , or, in other words, (¬φ, X (M/xn ), 1) ∈ T . Now, because of X = X (M/xn ) dom(X ) and Lemma 3.27, we get (¬φ, X, 1) ∈ T , whence (φ, X, 0) ∈ T . This completes the proof for φ ⇒∗ ∃xn φ. Now we prove (iv)(b). Let φ be a D-formula. We need to show ∀xn φ ⇒∗ φ. We could write a proof similar to that for (iv)(a), but instead we present a slightly more interesting (and shorter) proof that is based on the duality between ∃ and ∀. Namely, by (iv)(a), proved above, (iv)(a)

∀xn φ = ¬∃xn ¬φ ⇒ ¬¬φ ≡∗ φ and (iv)(a)

¬φ ⇒ ∃xn ¬φ ≡∗ ¬¬∃xn ¬φ = ¬∀xn φ. Therefore ∀xn φ ⇒ φ and ¬φ ⇒ ¬∀xn φ, so we have ∀xn φ ⇒∗ φ.

Exercise 3.28 We are completing the proof of Proposition 3.31. M is a model and we are in the middle of an induction on the complexity of a D-formula φ that is first order. Our induction hypothesis is that for all subformulas ψ of φ, it holds that (1) if (ψ, X, 1) ∈ T , then M |=s ψ for all s ∈ X ; (2) if (ψ, X, 0) ∈ T , then M |=s ¬ψ for all s ∈ X . We only have to complete the proof for the cases where φ is of the form ¬ψ and of the form ∃xn ψ. If φ is of form ¬ψ for some first order D-formula ψ, then we get from (¬ψ, X, 1) ∈ T that (ψ, X, 0) ∈ T . Here we can use our induction hypothesis (2) and get M |=s ¬ψ for all s ∈ X , which is exactly what we wanted. If φ is of form ¬ψ, then from (¬ψ, X, 0) ∈ T we get (ψ, X, 1) ∈ T . Here we can use our induction hypothesis (1) and get that for all s ∈ X , M |=s ψ, which is by first order logic the same as M |=s ¬¬ψ. If φ is of form ∃xn ψ for some first order D-formula ψ, then from (∃xn ψ, X, 1) ∈ T we get (ψ, X (F/xn ), 1) ∈ T for some F : X → M. The elements of X (F/xn ) are of form s(F(s)/xn ) for s ∈ X , so by induction hypothesis 1 we get for each s ∈ X that M |=s(F(s)/xn ) ψ, whence M |=s ∃xn ψ. If φ is of form ∃xn ψ, then (∃xn ψ, X, 0) ∈ T gives (ψ, X (M/xn ), 0) ∈ T . Elements of X (M/xn ) are of form s(a/xn ) for s ∈ X and a ∈ M, so by induction hypothesis (2) we get for each s ∈ X that M |=s(a/xn ) ¬ψ holds for each a ∈ M, whence M |=s ∀xn ¬ψ. By first order logic, ∀xn ¬ψ ≡ ¬¬∀xn ¬ψ ≡ ¬∃xn ψ, which completes the proof.

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Exercise 3.29 Before getting into the proper exercise, one might ask why in this exercise we ask for a first order formula that is only logically equivalent to the given D-formula and not strongly logically equivalent. The reason is simply that with first order formulas we always get strong logical equivalence for free if we have logical equivalence. Namely, assuming φ ≡ ψ and M |= X ¬φ, if M |= X ψ, then M |= X φ, which is a contradiction (when X = ∅). Now, because ψ is first order, M |= X ψ ∨ ¬ψ, from which we get M |= X ¬ψ. Thus we have showed ¬φ ⇒ ¬ψ. With a similar proof we see ¬ψ ⇒ ¬φ, so we get φ ≡∗ ψ. (i) We claim that ∃x0 (=(x1 , x0 ) ∧ P x0 ) ≡ ∃x0 P x0 . The following chain of equivalences is based on definitions, except for the equivalence marked with (∗): (∃x0 (=(x1 , x0 ) ∧ P x0 ), X, 1) ∈ T ⇐⇒ (=(x1 , x0 ), X (F/x0 ), 1) ∈ T and (P x0 , X (F/x0 ), 1) ∈ T for some F : X → M (∗)

⇐⇒ (P x0 , X (G/x0 ), 1) ∈ T for some G : X → M ⇐⇒ (∃x0 P x0 , X, 1) ∈ T . The implication to the right at (∗) is clear because we can choose G to be F. To show the implication to the left, first note that if X = ∅ then the claim is clear as the empty team is of any type. Otherwise take some s0 ∈ X and assume (P x0 , X (G/x0 ), 1) ∈ T for some G : X → M. This means in particular that G(s0 ) ∈ P M . Now construct F : X → M by letting F(s) = G(s0 ) for all s ∈ X . In other words, F resembles G but has the additional good property of being constant (it is good concerning determination and dependence). Now (P x0 , X (F/x0 ), 1) ∈ T because if s ∈ X (F/x0 ) then s(x0 ) = G(s0 ) ∈ P M . Also (=(x1 , x0 ), X (F/x0 ), 1) ∈ T because when s, s  ∈ X (F/x0 ), we have s(x0 ) = G(s0 ) = s  (x0 ). (ii) We claim that ∃x0 (=(x1 , x0 ) ∧ P x1 ) ≡ ∃x0 P x1 . The following chain of equivalences is based on definitions, except for the equivalence marked with (∗): (∃x0 (=(x1 , x0 ) ∧ P x1 ), X, 1) ∈ T ⇐⇒ (=(x1 , x0 ), X (F/x0 ), 1) ∈ T and (P x1 , X (F/x0 ), 1) ∈ T for some F : X → M (∗)

⇐⇒ (P x1 , X (G/x0 ), 1) ∈ T for some G : X → M ⇐⇒ (∃x0 P x1 , X, 1) ∈ T . The implication to the right at (∗) is clear because we can choose G to be F. To show the implication to the left, assume (P x1 , X (G/x0 ), 1) ∈ T for some G : X → M. Fix a ∈ M and construct F : X → M by letting F(s) = a for all s ∈ X . Then (P x1 , X (F/x0 ), 1) ∈ T because modifying values s(x0 ) for s ∈ X does not affect values s(x1 ). Also (=(x1 , x0 ), X (F/x0 ), 1) ∈ T because when s, s  ∈ X (F/x0 ), we have s(x0 ) = a = s  (x0 ). Note that we can further refine the result by noticing that in first order logic we have ∃x0 P x1 ≡ P x1 .

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  (iii) We claim that ∃x0 (=(x0 , x1 ) ∧ P x0 ) → P x1 ≡ ∃x0 (P x0 → P x1 ). First note that, by Lemma 3.25, Preservation of Equivalence under Substitution,     ∃x0 (φ ∧ ψ) → θ = ∃x0 ¬(φ ∧ ψ) ∨ θ ≡∗   ∃x0 ¬(¬¬φ ∧ ¬¬ψ) ∨ θ = ∃x0 (¬φ ∨ ¬ψ ∨ θ). Now we can deduce as follows (the explanation for (∗) is given in the following text): (∃x0 ((=(x1 , x0 ) ∧ P x0 ) → P x1 ), X, 1) ∈ T ⇐⇒ (∃x0 (¬ =(x1 , x0 ) ∨ ¬P x0 ∨ P x1 ), X, 1) ∈ T ⇐⇒ (¬ =(x1 , x0 ) ∨ ¬P x0 ∨ P x1 , X (F/x0 ), 1) ∈ T for some F : X →M ⇐⇒ (¬ =(x1 , x0 ), A, 1) ∈ T and (¬P x0 , B, 1) ∈ T and (P x1 , C, 1) ∈ T for some A ∪ B ∪ C = X (F/x0 ) and F : X → M (∗)

⇐⇒ (¬P x0 , B, 1) ∈ T and (P x1 , C, 1) ∈ T for some B ∪ C = X (F/x0 ) and F : X → M ⇐⇒ (¬P x0 ∨ P x1 , X (F/x0 ), 1) ∈ T for some F : X → M ⇐⇒ (∃x0 (P x0 → P x1 ), X, 1) ∈ T . The equivalence (∗) is based on the fact that (¬ =(x1 , x0 ), A, 1) ∈ T iff A = ∅. By first order logic we can further refine the formula by noting that ∃x0 (P x0 → P x1 ) ≡ ∃x0 (¬P x0) ∨ P x1 .  (iv) We claim that ∃x0 =(x1 , x0 ) ∧ Rx0 x1 ≡ ∃x0 Rx0 x1 . We can deduce (the explanation for (∗) can be found in the following text): (∃x0 (=(x1 , x0 ) ∧ Rx0 x1 ), X, 1) ∈ T ⇐⇒ (=(x1 , x0 ), X (F/x0 ), 1) ∈ T and (Rx0 x1 , X (F/x0 ), 1) ∈ T for some F : X → M (∗)

⇐⇒ (Rx0 x1 , X (F/x0 ), 1) ∈ T for some F : X → M ⇐⇒ (∃x0 Rx0 x1 , X, 1) ∈ T . The implication to the right at (∗) is clear. To justify the implication to the left at (∗), assume (Rx0 x1 , X (F/x0 ), 1) ∈ T for some F : X → M. We can assume that dom(X ) = {1} because Lemma 3.27 shows that values of variables that do not appear in the formula do not matter. Now, if we have s, s  ∈ X (F/x0 ) with s(x1 ) = s  (x1 ), they are of form s = z(F(z)/x0 ) and s  = z  (F(z  )/x0 ) for some z, z  ∈ X . Because dom(z) = {1} and dom(z  ) = {1}, and s(x1 ) = z(1) and s  (x1 ) = z  (1), we have z = z  . Thus F(z) = F(z  ), so s(x0 ) = F(z) = F(z  ) = s  (x0 ). Therefore (=(x1 , x0 ), X (F/x0 ), 1) ∈ T .

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  (v) We claim that ∃x0 =(x0 ) ∧ (Rx1 x0 ∨ Rx0 x0 ) ≡ ∃x0 (Rx1 x2 ∨ Rx0 x0 ). We can deduce as follows (explanation for (∗) can be found in the following text): (∃x0 (=(x0 ) ∧ (Rx1 x2 ∨ Rx0 x0 )), X, 1) ∈ T ⇐⇒ (=(x0 ), X (F/x0 ), 1) ∈ T and (Rx1 x2 ∨ Rx0 x0 , X (F/x0 ), 1) ∈ T for some F : X → M (∗)

⇐⇒ (Rx1 x2 ∨ Rx0 x0 , X (G/x0 ), 1) ∈ T for some G : X → M ⇐⇒ (∃x0 (Rx1 x2 ∨ Rx0 x0 ), X, 1) ∈ T . The implication to the right at (∗) is clear. To justify the implication to the left, assume (Rx1 x2 ∨ Rx0 x0 , X (G/x0 ), 1) ∈ T for some G : X → M. Then (Rx1 x2 , A, 1) ∈ T and (Rx0 x0 , B, 1) ∈ T for some A ∪ B = X (G/x0 ). In fact we have A = Y (G/x0 ) and B = Z (G/x0 ) for some Y ∪ Z = X . Now we examine two cases. The first case is when Z = ∅. Note that we have for all s ∈ Z : (G(s), G(s)) ∈ R M .

(A.2)

Now pick some s0 ∈ Z and define F : X → M by F(s) = G(s0 ) for all s ∈ X . Now, if s, s  ∈ X (F/x0 ), it holds that s(x0 ) = s  (x0 ) because F is a constant function. Therefore (=(x0 ), X (F/x0 ), 1) ∈ T . We also have (Rx1 x2 , ∅, 1) ∈ T because the empty team is of any type. Finally, we have (Rx0 x0 , X (F/x0 ), 1) ∈ T because if s ∈ X (F/x0 ), then s(x0 ) = G(s0 ), and by Eq. (A.2) and the fact that s0 ∈ Z , we get (s(x0 ), s(x0 )) ∈ R M . From these we gather (Rx1 x2 ∨ Rx0 x0 , ∅ ∪ X (F/x0 ), 1) ∈ T .  

X (F/x0 )

The second case is when Z = ∅. Then we have, for all s ∈ X , (s(x1 ), s(x2 )) ∈ R M . Let a ∈ M and define F : X → M by F(s) = a for all s ∈ X . Still we have, for all s ∈ X , (s(x1 ), s(x2 )) ∈ R M . Therefore (Rx1 x2 ∨ Rx0 x0 , X (F/x0 ) ∪ ∅, 1) ∈ T .  

X (F/x0 )

As in the first case, we have (=(x0 ), X (F/x0 ), 1) ∈ T because F is a constant function, and this completes the proof.

Exercise 3.31 (i) The D-formula =() ∨ ¬ =() is in itself first order, by definition, because =() is the veritas symbol, . (ii) The D-formula =(x0 ) is not logically equivalent to a first order formula because it fails the Flatness Test. Let M = {a, b} have two different elements, s = {(0, a)} and s  = {(0, b)}. Then M |={s} =(x0 ) and M |={s  } =(x0 ) but M |={s,s  } =(x0 ). (iii) We claim that =(x0 , x0 ) ≡ . We can prove it by noting that M |= X =(x0 , x0 ) ⇐⇒ for all s, s  ∈ X ; it holds that if s(x0 ) = s  (x0 ) then s(x0 ) = s  (x0 ). This is of course always true, as is M |= X .

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Exercise 3.38 For each of the three given formulas we must find a model and a team that demonstrate that the formula fails the Flatness Test.  (i) Let φ be ∃x0 =(x2 , x0 ) ∧ Rx0 x1 . Let M = (Z, <), X = {sk : k ∈ Z} ∈ Team(Z, {x1 , x2 }), where sk (x1 ) = k and sk (x2 ) = 0 for each k ∈ Z. We have   for each sk ∈ X that φ, {sk }, 1 ∈ T because =(x2 , x0 ) ∧ Rx0 x1 , {sk }(F/x0 ), 1 ∈ T  , where F : sk → k− 1. But (φ, X, 1) ∈ T because if G : X → M satisfies =(x2 , x0 ), X (G/x0 ), 1 ∈ T then G must be a constant function because X has a constant value for x2 , namely 0. That is, there  is some n ∈ Z with  G(sk ) = n for all sk ∈ X . But now G(sn ) = n <n = sn (x1 ), so Rx0 x1 , X (G/x  0 ), 1 ∈ T . (ii) Let ψ be ∃x0 =(x2 , x0 ) ∧ Rx1 x0 ∧ Rx2 x0 . Let M = (N, <), X = {sk : k ∈ N},  where sk are as above. Now we have, for all sk ∈ X , ψ, {sk }, 1 ∈ T because =(x2 , x0 ) ∧ Rx1 x0 ∧ Rx2 x0 , {sk }(F/x0 ), 1 ∈ T , where F : sk → k + 1. However,  (ψ, X, 1) ∈ T because, as before, if G : X → M satisfies =(x2 , x0 ), X (G/x0 ), 1 ∈ T , then G must be a constant function,mapping G(sk ) = nfor all sk ∈ X and a fixed n ∈ N. Then sn (x1 ) = n < n = G(sn ), so Rx1 x0 , X (G/x0 ),1 ∈ T . (iii) Let θ be ∃x0 =(x2 , x0 ) ∧ (Rx1 x0 ↔ Rx2 x0 ) . Let M = (Z, <), X = {sk : k ∈ Z},  where sk are as above. Now, for each sk ∈ X , we have θ, {sk }, 1 ∈ T because =(x2 , x0 ) ∧ (Rx1x0 ↔ Rx2 x0 ), {sk }(F/x0 ), 1 ∈ T , where F : sk → max{k + 1, 1}.  However, θ, X, 1 ∈ T because if G : X → M satisfies =(x2 , x0 ), X (G/x0 ), 1 ∈ T , G has to be a constant function, as before, mapping G(sk ) = n for all sk ∈ X and a fixed n ∈ Z. If n ≤ 0 then sn−1 (x1 ) = n − 1 < n = G(sn−1 ) but sn−1 (x2 ) = 0 < n = G(sn−1 ). If n> 0 then sn (x1 ) = n < n = G(s  n ) but sn (x2 ) = 0 < n = G(sn ). In either case we get Rx1 x0 ↔ Rx2 x0 , X (G/x0 ), 1 ∈ T .

Exercise 3.40 In this exercise, we call a set P = {s, s  } for s, s  ∈ X a pair. Note that a pair P may have one or two elements.   Let ψ be ∃x0 =(x1 , x0 ) ∧ φ , where φ is first order, and Fr(φ) = {x1 }. The goal is to show that ψ is coherent, i.e. M |= X ψ ⇐⇒ for all pairs P: M |= P ψ.

(A.3)

The implication to the right in Eq. (A.3) is clear by the Closure Test. Then we prove Eq. (A.3) from right to left. By Lemma 3.27 and the assumption that Fr(φ) = {x1 }, it is sufficient to examine only the case when dom(X ) = {1}. So, assume that for each pair P we have ψ, P, 1) ∈ T . That is, for each pair P we have FP : P → M with 

 =(x1 , x0 ) ∧ φ, P(FP /x0 ), 1 ∈ T .

(A.4)

 Define F : X → M by letting  F(s) = F{s} (s) for each s ∈ X. We show =(x1 , x0 ), X (F/x0 ), 1 ∈ T . Let s, s ∈ X (F/x0 ) and assume s(x1 ) = s (x1 ). We can write s = z(F(z)/x0 ) and s  = z  (F(z  )/x0 ) for some z, z  ∈ X . Then z(1) = z  (1), so because dom(X ) = {1}, z = z  and s(x0 ) = F(z) = F(z  ) = s  (x0 ). Then we show φ, X (F/x0 ), 1 ∈ T . Let s ∈ X (F/x0 ). Then s = z(F(z)/x0 ) = z(F{z} (z)/x0 ) for some z ∈ X , so {s} = {z}(F{z} /x0 ). Now, by Eq. (A.4), we get

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Table A.5. The teams X and X (F/x0 ) for the Exercise 3.40 counterexample, x ∈ {a, b, c}. X x0 s1 s2 s3

X (F/ X 0 ) x1

x2

a a a

a b c

s1 s2 s3

x0

x1

x2

x x x

a a a

a b c

  φ, {s}, 1 ∈ T . Because φ is first order, it passes the Flatness Test. Therefore φ, X (F/x0 ), 1 ∈ T .   By combining these achievements, we get =(x1 , x0 ) ∧ φ, X (F/x0 ), 1 ∈ T , whence (ψ, X, 1) ∈ T . This completes the proof that ψ is coherent. Note that if we drop the requirement that Fr(φ) = {x1 }, the claim does not hold. We can then choose a counterexample by letting φ be ¬(x0 = x2 ), letting M = {a, b, c} have three different elements, and letting X = {sz : z ∈ M}, where sz = {(x1 , a), (x2 , z)} for each z ∈ M (see Table A.5). We have, for each P, M |= P ∃x0 (=(x1 , x0 ) ∧ φ) because for each pair P there is some z ∈ M with sz ∈ P, so we can choose FP : P → M  to map F (s) = z. Then =(x , x0 ), P(FP /x0 ), 1) ∈ T because FP is constant, and P 1  φ, P(FP /x0 ), 1) ∈ T . But we do not have M |=X ∃x0 (=(x1 , x0 ) ∧ φ) because any  F : X → M satisfying =(x1 , x0 ) ∧ φ, X (F/x0 ), 1 ∈ T would have to be constant; i.e. for some z ∈ M we had F(s) = z for all s ∈ X . But then sz (2) = F(sz ), whence (φ, X (F/x0 ), 1) ∈ T .

Exercise 3.41 Before we get to a straight solution of the exercise, here is an analogy that may help the student to solve the problem alone. Compare the notions “φ passes the Flatness Test” and “φ is coherent.” Flatness means M |= X φ ⇐⇒ ∀s ∈ X : M |={s} φ, whereas coherence means M |= X φ ⇐⇒ ∀s, s  ∈ X : M |={s,s  } φ. The similarity is clear. The only difference between the two concepts is that one considers singleton teams and the other considers pair teams. Now we can find an answer to this exercise by recalling a typical non-flat formula, =(x0 ), and the fact that disjunction splits teams into smaller ones, possibly into singletons or pairs. To solve the exercise, let φ be the formula =(x0 ) ∨ =(x0 ). We claim that φ is not coherent. Let M = {a, notation  b, c} have threedifferent elements, and let X = M (shorthand  meaning X = {(0, x)} : x ∈ M ). Whens, s  ∈ X , we have φ, {s, s }, 1) ∈ T because   we can split {s, s  } = {s} ∪ {s  } and have =(x0 ), {s}, 1 ∈ T and =(x0 ), {s  }, 1 ∈ T . But we do not have (φ, X, 1) ∈ T because whenever we split X = Y ∪ Z , either Y or Z has two different elements that do not agree on x0 . As an afterthought, note that =(x0 ), while being non-flat, is however coherent. Namely, if M |={s,s  } =(x0 ) for all s, s  ∈ X , then X is constant in x0 . Then M |= X =(x0 ).

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Exercise 3.43 Assume M |= X φ → ψ and M |= X ¬ψ. From the first assumption we get M |= X ¬φ ∨ ψ, whence M |=Y ¬φ and M |= Z ψ for some Y ∪ Z = X . Now Z ⊆ X , so by the Closure Test and the second assumption we also have M |= Z ¬ψ. We can apply Corollary 3.41 to the fact that now M |= Z ψ ∧ ¬ψ, and we get Z = ∅. Hence Y = X , which means that M |= X ¬φ, as we wanted.

Exercise 3.44 A straightforward way to show that no non-empty team is of a given type φ is to apply the truth definition to the formula and see what conditions it poses on the team. There is also another way to check this, namely to flatten the formula and then, remembering that φ ⇒ φ f , check for the satisfiability of the resulting first order formula φ f . The flattening technique is sometimes easier, especially if the formula involves determination atoms =(t1 , . . . , tn ). f  (i) ¬ =(x0 , x1 ) = ¬, which is clearly not satisfiable in first order logic. Therefore no non-empty  team is of type ¬ =(x0 , x1 ). (ii) Let φ be ¬ =(x0 , x1 ) → =(x2 , x1 ) . Then φ f = ¬( → ) ≡ ¬. As ⊥ is not satisfiable in first order logic, no non-empty team can be of type φ. (iii) Let ψ be ¬ =( f x0 , x0 ) ∨ ¬ =(x0 , f x0 ). Then φ f = ¬ ∨ ¬ = ⊥ ∨ ⊥ ≡ ⊥. As before, this is not satisfiable,  so no non-empty  team can be of type ψ. (iv) Let θ be ∀x0 ∃x1 ∀x2 ∃x3 ¬ φ → =(x0 , x1 ) , where φ is an arbitrary D-formula. Then θ f = ∀x0 ∃x1 ∀x2 ∃x3 ¬(φ f → ) ≡ ∀x0 ∃x1 ∀x2 ∃x3 ¬ ≡ ∀x0 ∃x1 ∀x2 ∃x3 ⊥ ≡ ⊥. Therefore no non-empty team can be of type θ.

Exercise 3.45 First, M |= X =(x0 , x2 ) ∧ =(x1 , x2 ) iff for all s, s  ∈ X it holds that if s(x0 ) = s  (x0 ) then s(x2 ) = s  (x2 ), and if s(x1 ) = s  (x1 ) then s(x2 ) = s  (x2 ). In other words, for all s, s  ∈ X it holds that if s(x0 ) = s  (x0 ) or s(x1 ) = s  (x1 ), then s(x2 ) = s  (x2 ). On the other hand, M |= X =(x0 , x1 , x2 ) iff for all s, s  ∈ X it holds that if s(x0 ) = s  (x0 ) and s(x1 ) = s  (x1 ), then s(x2 ) = s  (x2 ). Looking at these two observations we can see that the latter is a weaker claim, i.e. =(x0 , x1 , x2 ) ⇒ =(x0 , x2 ) ∧ =(x1 , x2 ), but the converse consequence does not hold. The team on the left in Table A.6 is of type =(x0 , x2 ) ∧ =(x1 , x2 ) because we have a function f mapping the values of the feature x0 to the values of the feature x2 , and we also have a function g mapping the values of the feature x1 to the values of the feature x2 . These two mappings are f : {1, 2} → {1, 2}, f (2) = 1, f (1) = 2, and g : {1, 2} → {1, 2, 3, 4}, g(1) = 1, g(2) = 1, g(3) = 2, g(4) = 2. The team on the right in Table A.6 is of type =(x0 , x1 , x2 ) because we have a function h mapping the value pairs of the features x0 and x1 to the values of the feature x2 . This mapping is h : {(3, 1), (4, 2), (4, 1), (3, 2)} → {1, 2}, h(3, 1) = 1, h(4, 2) = 1, h(4, 1) = 2, h(3, 2) = 2. The team on the right in Table A.6 is not of type =(x0 , x2 ) ∧ =(x1 , x2 ) because, for example, there cannot be a mapping f  : {3, 4} → {1, 2} that would map f  (3) = 1, f  (4) = 1, f  (4) = 2, f  (3) = 2.

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Table A.6. Two example teams for Exercise 3.45, one of type =(x0 , x2 ) ∧ =(x1 , x2 ) and another of type =(x0 , x1 , x2 ) =(x0 , x2 ) ∧ =(x1 , x2 )

=(x0 , x1 , x2 )

x0

x1

x2

x0

x1

x2

2 2 1 1

1 2 3 4

1 1 2 2

3 4 4 3

1 2 1 2

1 1 2 2

Exercise 3.48

 Assume φ is a first order D-formula. Denote by ψ the formula =(x0 , x1 ) ∧ =(x2 , x3 ) ∧  φ . We want to show that ∀x0 ∃x1 ∀x2 ∃x3 ψ ≡ ∀x2 ∃x3 ∀x0 ∃x1 ψ. We aim to establish the following chain of equivalences: ∀x0 ∃x1 ∀x2 ∃x3 ψ ≡ ∀x0 ∀x2 ∃x1 ∃x3 ψ

(A.5)

≡ ∀x2 ∀x0 ∃x3 ∃x1 ψ

(A.6)

≡ ∀x2 ∃x3 ∀x0 ∃x1 ψ.

(A.7)

The equivalences in Eqs. (A.5) and (A.7) require a long explanation. The equivalence in Eq. (A.6) is obtained by utilizing Lemma 3.23 twice along with the Preservation of Equivalence under Substitution. Let us now immerse ourselves into Eqs. (A.5) and (A.7). To deal with some ample notation in solving this exercise, we improve the shorthand notation used in the book. Let M be a model, let X be a team, and let s ∈ X . We denote s(a/xm )(b/xn ) by s(ab/xm xn ) for a, b ∈ M. To obtain the equivalence in Eq. (A.5), we need to show, for any model M and team X , that (∀x0 ∃x1 ∀x2 ∃x3 ψ, X, 1) ∈ T ⇐⇒ (∀x0 ∀x2 ∃x1 ∃x3 ψ, X, 1) ∈ T . By applying the truth definition we get that (∀x0 ∃x1 ∀x2 ∃x3 ψ, X, 1) ∈ T if and only if (ψ, Y, 1) ∈ T , where Y = X (M/x0 )(F/x1 )(M/x2 )(G/x3 ) for some functions F : X (M/x0 ) → M and G : X (M/x0 )(F/x1 )(M/x2 ) → M. This is equivalent to the following condition: (=(x0 , x1 ), Y, 1) ∈ T and (=(x2 , x3 ), Y, 1) ∈ T and (φ, Y, 1) ∈ T .

(A.8)

We are heading towards (∀x0 ∀x2 ∃x1 ∃x3 ψ, X, 1) ∈ T . Let us unravel it slightly by the truth definition. It is equivalent to (ψ, Y  , 1) ∈ T , where Y  = X (M/x0 )(M/x2 )(F  /x1 )(G  /x3 ) for some functions F  : X (M/x0 )(M/x2 ) → M and G  : X (M/x0 )(M/x2 )(F  /x1 ) → M. This is equivalent to the following condition: (=(x0 , x1 ), Y  , 1) ∈ T and (=(x2 , x3 ), Y  , 1) ∈ T and (φ, Y  , 1) ∈ T .

(A.9)

In order to prove Eq. (A.5), we need to show that Eqs. (A.8) and (A.9) are equivalent.

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Let us first prove that Eq. (A.8) implies Eq. (A.9). We first need to define F  and G  . Let z(ac/x0 x2 ) ∈ X (M/x0 )(M/x2 ), where z ∈ X and a, c ∈ M. We of course want to use the given function F when we define F  . A natural way to do this is to set F  (z(ac/x0 x2 )) = F(z(a/x0 )). We can do so because z(a/x0 ) is in the domain of F. Let z(acb/x0 x2 x1 ) ∈ X (M/x0 )(M/x2 )(F  /x1 ), where z ∈ X , a, c ∈ M, and b = F  (z(ac/x0 x2 )). Again we do the natural thing and set G  (z(acb/x0 x2 x1 )) = G(z(abc/x0 x1 x2 )). We can do this because b = F  (z(ac/x0 x2 )) = F(z(a/x0 )), so z(abc/x0 x1 x2 ) is in the domain of G. In fact, z(acb/x0 x2 x1 ) and z(abc/x0 x1 x2 ) are the same assignment, so G  and G are the same function, but we are not going to pay attention to this. Now we check that F  and G  satisfy Eq. (A.9). Consider first =(x0 , x1 ). Let s1 = z 1 (a1 c1 b1 d1 /x0 x2 x1 x3 ) ∈ Y  and s2 = z 2 (a2 c2 b2 d2 /x0 x2 x1 x3 ) ∈ Y  , where, for i = 1, 2, z i ∈ X , ai , ci ∈ M, bi = F  (z i (ai ci /x0 x2 )), and di = G  (z i (ai ci bi /x0 x2 x1 )). Recall that bi = F  (z i (ai ci /x0 x2 )) = F(z i (ai /x0 )) and di = G  (z i (ai ci bi /x0 x1 x2 )) = G(z i (ai bi ci /x0 x1 x2 )). Assume s1 (x0 ) = s2 (x0 ), i.e. a1 = a2 . Then 

 z 1 (a1 b1 /x0 x1 ) (x0 ) = a1 = a2 = z 2 (a2 b2 /x0 x1 ) (x0 ).

Now, because from Eq. (A.8) we get (=(x0 , x1 ), Y, 1) ∈ T , and because Lemma 3.27 says that satisfaction is unaffected by assignments to variables that do not occur, and because we can extend the assignments z i (ai bi /x0 x1 ) to some such assignments that are elements of Y , we find   b1 = z 1 (a1 b1 /x0 x1 ) (1) = z 2 (a2 b2 /x0 x1 ) (1) = b2 .   Therefore s1 (x1 ) = b1 = b2 = s2 (x1 ), which proves that =(x0 , x1 ), Y  , 1 ∈ T . Consider =(x2 , x3 ). Let s1 , s2 ∈ Y  as above. Assume s1 (x2 ) = s2 (x2 ), i.e. c1 = c2 . Then 

 z 1 (a1 b1 c1 d1 /x0 x1 x2 x3 ) (x2 ) = c1 = c2 = z 2 (a2 b2 c2 d2 /x0 x1 x2 x3 ) (x2 ).

As above, because from Eq. (A.8) we get (=(x2 , x3 ), Y, 1) ∈ T , and because we have Lemma 3.27, and because we can extend the assignments z i (ai bi ci di /x0 x1 x2 x3 ) to some such assignments that are elements of Y , we find   d1 = z 1 (a1 b1 c1 d1 /x0 x1 x2 x3 ) (x3 ) = z 2 (a2 b2 c2 d2 /x0 x1 x2 x3 ) (x3 ) = d2 .   Therefore s1 (x3 ) = d1 = d2 = s2 (x3 ), which proves that =(x2 , x3 ), Y  , 1 ∈ T . Finally, consider φ. Because φ is first order, we can use the Flatness Test. To show (φ, Y  , 1) ∈ T it suffices to show that (φ, {s}, 1) ∈ T for each s ∈ Y  . Let s = z(acbd/x0 x2 x1 x3 ) ∈ Y  as above. Because b = F  (z(ac/x0 x2 )) = F(z(a/x0 )) and d = G  (z(acb/x0 x2 x1 )) = G(z(abc/x0 x1 x2 )), we have s ∈ Y . Therefore (φ, {s}, 1) ∈ T , by Eq. (A.8) and the Closure Test. This concludes the proof that Eq. (A.8) implies Eq. (A.9). The other direction is similar. Also the proof of Eq. (A.7) is similar to the proof of Eq. (A.5).

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Chapter 4 Exercise 4.1 In order to come up with a D-sentence that would characterize the oddness of the universe, we can start from the mathematical expression and then modify it into equivalent forms using an informal second order language. We know that  in D we can  describe the existence of a function by writing something like ∀x∃y =(x, y) ∧ φ . Here x is the argument for the function, y is the value the function gives to the argument, and φ describes any conditions that we want to pose on the function. So we begin. We proceed from the mathematical expression Eq. (A.10) via the equivalent intermediate conditions Eqs. (A.11) and (A.12) to the D-sentence Eq. (A.13). The basic idea for characterizing oddness is similar to that for evenness: a finite set M is even iff there is a one-to-one function f : M → M that maps no element to itself and that is its own inverse. Here are the three equivalent conditions: |M| is odd;

(A.10)

there is an element x ∈ M and a function f : M → M such that f ◦ f = id and for all y ∈ M we have f (y) = y iff y = x;

    ∃x∃ f ∀y f ( f (y)) = y ∧ ∀y f (y) = y ↔ y = x .

(A.11) (A.12)

Now we bring in the D-way of talking about functions. In Eq. (A.13) (see below), the variable x1 acts as an argument to the function and the variable x2 carries the value that the function gives to its argument. Thus we may informally write y for x1 and f (y) for x2 . Actually we need to specify the function f twice because we need to be able to speak about mapping two different elements with it at the same time. Thus we think of x3 as an argument to f and x4 as its image. We can informally write z for x3 and f (z) for x4 . Now, in order to express f ( f (y)) we can and must express it as “ f (z) when z = f (y).” Using this intuition we can note how the subformula x3 = x2 → x4 = x1 in Eq. (A.13) expresses that x3 = f (x1 ) → f (x3 ) = x1 , i.e. f ( f (x1 )) = x1 . Also note how x1 = x3 → x2 = x4 expresses that x = y → f (x) = g(y), i.e. f (x) = g(x), i.e. f = g. So here is our final D-sentence: ∃f

∃g

x2 = f (x1 )

x4 =g(x3 )

f =g

 

         ∃x0 ∀x1 ∃x2 ∀x3 ∃x4 =(x1 , x2 ) ∧ =(x3 , x4 ) ∧ (x1 = x3 → x2 = x4 ) ∧

 (x3 = x2 → x4 = x1 ) ∧ (x2 = x1 ↔ x1 = x0 ) . 



  f ( f (y))=y

(A.13)

f (x1 )=x1 ↔x1 =x0

Exercise 4.15 Let G = (V, E) be a graph, let V be its vertices, and let E be its edges (a binary relation). We proceed as in Exercise 4.1, starting from the mathematical expression, proceeding to equivalent expressions, and ending up with a D-sentence that describes the mathematical

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expression. Here are the two equivalent conditions: G has an infinite clique;

(A.14)

there is a subset A ⊆ V such that A is a clique and A is infinite.

(A.15)

We can describe infinity of A by saying that there is a one-to-one function f : A → A that is not onto

∃A ∀x ∈ A∀y ∈ A(x E y) ∧ ∃ f : A → A ∀x ∈ A∀y ∈ A     f (x) = f (y) → x = y ∧ ∃z ∈ A∀x ∈ A f (x) = z .

(A.16)

We need to speak about subsets of the universe. A way to specify a subset, using functions as the only tool, is to speak of the subset as the pre-image of a fixed element in a fixed function. Therefore we convert ∃A into ∃g∃w and expressions such as x ∈ A into f (x) = w. We also allow f to map M → M, and we require it to be one-to-one only for elements x, y ∈ A:

  ∃g∃w ∀x∀y (g(x) = w ∧ g(y) = w) → x E y

   ∧ ∃ f ∀x∀y g(x) = w ∧ g(y) = w ∧ f (x) = f (y) → x = y

  ∧ ∃z g(z) = w ∧ ∀x g(x) = w → f (x) = z .

(A.17)

Now we convert our informal and vaguely defined second order sentence into a D-sentence. Note that in order to ensure that w corresponds to a single value, we must express it by =(w), which says that w is a constant. If we leave it out, w could have different values for different arguments of function g, i.e. w would be another function. As in Exercise 4.1, we can think of x0 as x and x1 as its value, g(x). Likewise, think of x2 as y and x3 as g(y). So, finally, the D-sentence is given by ∃g

∃g 

∃w

x1 =g(x0 )

x3 =g  (x2 )

x4 constant

           ∀x0 ∃x1 ∀x2 ∃x3 ∃x4 =(x0 , x1 ) ∧ =(x2 , x3 ) ∧ =(x4 ) g=g 

g(x)=w ∧ g(y)=w

x Ey

  

      ∧ x0 = x2 → x1 = x3 ∧ (x1 = x4 ∧ x3 = x4 ) → x0 E x2 ∃f

∃f

x6 = f (x5 )

x8 = f  (x7 )

        ∧ ∀x5 ∃x6 ∀x7 ∃x8 =(x5 , x6 ) ∧ =(x7 , x8 ) g(x)=w ∧ g(y)=w

g and f map same values

f (x)= f (y)

x=y

 

        ∧ (x1 = x4 ∧ x3 = x4 ∧ x0 = x5 ∧ x2 = x7 ∧ x6 = x8 ) → x0 = x2 ∃z

 ∧ ∃x9 =(x9 ) ∧ x2 = x9 ∧ x3 = x4 ∧   (x1 = x4 ∧ x5 = x0 ) → (x6 = x9 ) .

(A.18)

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Exercise 4.18 Let G = (V, E) be a graph. We proceed from Eq. (A.19) via equivalent, informal, second order expressions to the D-formula Eq. (A.23). We see that the following are equivalent: G is 3-colorable.

(A.19)

G can be partitioned into three subsets, A0 , A1 , A2 , such that no edge goes from one subset to the same subset;

 Ai ∩ A j = ∅ ∧ A 0 ∪ A 1 ∪ A 2 = V ∃A0 ⊆ V ∃A1 ⊆ V ∃A2 ⊆ V

(A.20)

i< j<3

   ∧ ∀x∀y x E y → ¬(x ∈ Ai ∧ y ∈ Ai ) .

(A.21)

i<3

As in Exercise 4.15, we express each subset Ai as the pre-image of a fixed element, f i−1 {z}. As an aside, we could save some quantifiers and take just one function and three elements and write f −1 {z i } for Ai . We do not do this, however, because then the resulting formula would work only in models with three different elements. That would not be a real problem, because we can then write a separate formula that would describe 3-colorability for the cases when the universe has less than three elements. But as it might look unnecessarily complicated and it would just save us a few quantifiers, we do not do it now:

    ∃ f 0 ∃ f 1 ∃x2 ∃z ¬∃x f i (x) = z ∧ f j (x) = z ∧ ∀x f i (x) = z i< j<3

i<3

   ∧ ∀x∀y x E y → ¬ f i (x) = z ∧ f i (y) = z . 

(A.22)

i<3

The corresponding D-sentence is given by

∀x0 ∃x1 ∀x2 ∃x3 ∀x4 ∃x5 ∃x6 =(x0 , x1 ) ∧ =(x2 , x3 ) ∧ =(x4 , x5 ) ∧ =(x6 )    ∧ x2i = x2 j → ¬(x2i+1 = x6 ∧ x2 j+1 = x6 ) i< j<3

  ∧ (x0 = x2 ∧ x2 = x4 ) → (x1 = x6 ∨ x3 = x6 ∨ x5 = x6 ) ∧ ∀x7 ∀x8

   ¬ (x0 = x7 → x1 = x6 ) ∧ (x0 = x8 → x1 = x6 ) . (A.23) x7 E x8 → i<3

Chapter 6 Exercise 6.1 Because ∃x1 =(x1 ) is a sentence, we can omit the relation symbol S from τd,∃x1 =(x1 ) for both d = 0, 1: τ1,∃x1 =(x1 )   = ∃R τ1,=(x1 ) (R) ∧ ∃x1 Rx1  = ∃R ∀x1 ∀x2 ((Rx1 ∧ Rx2 )  → x1 = x2 ) ∧ ∃x1 Rx1 ;

τ0,∃x1 =(x1 )   = ∃R τ0,=(x1 ) (R) ∧ ∀x1 Rx1   = ∃R ∀x1 ¬Rx1 ∧ ∀x1 Rx1 .

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Exercise 6.2 τ1,∃x1 (=(x1 )∨x1 =x0 ) (S)   = ∃R τ1,=(x1 )∨x1 =x0 (R) ∧ ∀x0 (Sx0 → ∃x1 Rx0 x1 )

= ∃R∃T1 ∃T2 τ1,=(x1 ) (T1 ) ∧ τ1,∨x1 =x0 (T2 ) ∧    ∀x0 ∀x1 Rx0 x1 → (T1 x1 ∨ T2 x0 x1 ) ∧ ∀x0 (Sx0 → ∃x1 Rx0 x1 )

= ∃R∃T1 ∃T2 ∀x1 ∀x2 (T1 x1 ∧ T1 x2 →x1 = x2 ) ∧ ∀x0 ∀x1 (T2 x0 x1 →x1 = x0 ) ∧    ∀x0 ∀x1 Rx0 x1 → (T1 x1 ∨ T2 x0 x1 ) ∧ ∀x0 (Sx0 → ∃x1 Rx0 x1 ) .

Exercise 6.5 Assume φ(xi1 , . . . , xin ) is of form ∃xin+1 ψ(xi1 , . . . , xin+1 ). Our induction hypothesis is that the claim holds for the subformula, namely that (ψ, X, d) ∈ T iff (M, X ) |= τd,ψ (S). Recall the following definitions from Theorem 6.2:   τ1,φ (S) = ∃R τ1,ψ (R) ∧ ∀xi1 . . . ∀xin (Sxi1 . . . xin → ∃xin+1 Rxi1 . . . xin+1 ) ;   τ0,φ (S) = ∃R τ0,ψ (R) ∧ ∀xi1 . . . ∀xin (Sxi1 . . . xin → ∀xin+1 Rxi1 . . . xin+1 ) . By the truth definition of D, (φ, X, 1) ∈ T is equivalent with the pair (ψ, X (F/xin+1 ), 1) being in T for some F : X → M. By our induction hypothesis this is equivalent with (M, X (F/xin+1 )) |= τ1,ψ (S).

(A.24)

When we choose X (F/xin+1 ) to witness R, we see that Eq. (A.24) implies (M, X ) |= τ1,φ (S) . Namely, letting N be the model with universe N = M and interpretations S N = X , R N = X (F/xin+1 ), we get N |= τ1,ψ (R) and N |= ∀xi1 . . . ∀xin (Sxi1 . . . xin → ∃xin+1 Rxi1 . . . xin+1 ). The latter comes from the fact that F gives us a suitable value at ∃xin+1 . On the other hand, assuming (M, X ) |= τ1,φ (S) we get some model N with universe N = M and relation symbols R, S such that S N = X , N |= τ1,φ (R) and N |= ∀xi1 . . . ∀xin (Sxi1 . . . xin → ∃xin+1 Rxi1 . . . xin+1 ). The latter gives us a function F : X → M such that R N = X (F/xin+1 ), and (A.24) holds. Similarly for d = 0.

Exercise 6.6 We solve the exercise by proving a more general claim. The general claim is that when φ is a 11 -formula, M |=s φ for some assignment s, and π is an isomorphism M ∼ = N, then N |=π◦s φ. We prove this by induction on φ. Our induction hypothesis is that the claim holds for all subformulas of φ. As the first step of the proof, we can point out that if φ is first order, then it is well known that the claim holds. Assume then that φ is of form ∃Rψ. Now M |=s φ implies that there is a relation S such that (M, S) |=s ψ. By π : M ∼ = N we get (N , S  ) |=π◦s ψ, where S  is {(π (x1 ), . . . , π(xn )) : (x1 , . . . , xn ) ∈ S}, the image of S under the isomorphism π. By the truth definition of 11 , we get N |=π ◦s φ.

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Assume finally that φ is of form ∃gψ. Now M |=s φ implies that there is a function h : M n → M such that (M, h) |=s ψ. By π : M ∼ = N we get (N , h  ) |=π◦s ψ, where   n  h is h : N → N , h (π (x1 ), . . . , π(xn )) = π (h(x1 , . . . , xn )) for all x1 , . . . , xn ∈ M, the image of h under π. By the truth definition of 11 , we get N |=π◦s φ.

Exercise 6.12 Let Mn |= φn for each n < ω. From φn+1 ⇒ φn for all n < ω we can see with a small inductive proof that φn ⇒ φk for all k ≤ n < ω.

(A.25)

We use the Compactness Theorem. Let T be some finite set of sentences φn . Then there is a greatest k such that φk ∈ T . We have Mk |= φk , and by Eq. (A.25) this implies Mk |= φn for all φn ∈ T . Now we have a model for an arbitrary finite set of sentences φn . By the Compactness Theorem of D there is a model M such that M |= φn for all n < ω.

Exercise 6.13 Let M be some infinite model of φ. Assume towards contradiction that M |= ψ. Then M |= ¬ψ (remember that ψ is first order so negation works in this complementing way). As M |= φ ∧ ¬ψ, by the L¨owenheim–Skolem Theorem of D there is a countable model N of φ ∧ ¬ψ. Now we have N |= φ, and by our assumption we get also N |= ψ. But this is in contradiction with N |= ¬ψ. Therefore we must have M |= ψ. Hereby all models of φ are models of ψ.

Exercise 6.14 We use the vocabulary {<}, where < is a binary relation symbol. Let θLO be a first order sentence that lists the axioms of linear order. Let ψ be the first order sentence   θLO ∧ ∀x∀y x < y → ∃z(x < z ∧ z < y) ∧ ∃x∃y(x < y). What ψ says is that the model is a dense linear order with at least two elements. Let φ be the D-sentence ψ ∧ cmpl . Recall from Section 4.3 that cmpl is true in a linear order iff the linear order is incomplete. Clearly we have φ ⇒ ψ. To show that M |= ψ implies M |= φ for countable models M, note first that M |= ψ implies that M is an infinite model. We know that there are, up to isomorphism, only four countable dense linear orders, namely the rationals with or without endpoints: Q, Q + 1, 1 + Q, 1 + Q + 1. All of these are incomplete, so we get M |= φ. However, there are uncountable models that satisfy ψ but not φ, for example the reals, (R, <).

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Exercise 6.15 Here are Skolem normal forms for the three given first order sentences. There are also other solutions. Here f 0 is a 0-ary function symbol, so it works as a constant: ∃ f 1 ∀x0 (x0 = f 1 x0 ), ∃ f 0 ∀x1 ¬( f 0 = x1 ), ∃ f 0 ∀x1 (P f 0 ∨ ¬P x1 ).

Exercise 6.17 These are the Skolem Normal Forms for the two given 11 -sentences, obtained by the process described in the proof of the Skolem Normal Form Theorem:   (i) ∃ f 0 ∃ f ∀x1 ∀x2 ¬ f x1 = f 0 ∧ ( f f 0 = f x1 → f 0 = x1 ) ; (ii) ∃ f ∃ f 0 ∃ f 1 ∀x0 ∀x1 ∀x2 ∀x3 ∀x4 ∀x5 ∀x6   ( f 0 x0 x1 = f 1 ∧ f 0 x1 x2 = f 1 ) → f 0 x0 x2 = f 1

 ∧ ( f 0 x3 x4 = f 1 ∨ f 0 x4 x3 = f 1 ∨ x3 = x4 ) ∧ ¬( f 0 x5 x5 = f 1 ) ∧ f 0 x6 f x6 = f 1 .

Exercise 6.18 Since the three given 11 -sentences are in Skolem Normal Form and each function symbol appears in them with only one kind of list of arguments, we may straightforwardly transform them into D, following the proof of Theorem 6.15:   (i) ∀x0 ∀x1 ∃x2 ∃x3 =(x0 , x1 , x2 ) ∧ =(x0 , x1 , x3 ) ∧ φ(x0 , x1 , x2 ,x3 ) ; (ii) ∀x0 ∀x1 ∃x2∃x3 =(x0 , x1 , x2 ) ∧ =(x1 , x3 ) ∧ φ(x0 ,x1 , x2 , x3 ) ; (iii) ∀x0 ∃x1 ∃x2 =(x0 , x1 ) ∧ =(x0 , x2 ) ∧ φ(x0 , x1 , x2 ) .

Exercise 6.19 The function symbol f appears in the given 11 -sentence with two different lists of arguments. We must therefore first transform the sentence into the following (or similar) equivalent form:   ∃ f ∃ f  ∀x0 ∀x1 φ(x0 , x1 , f x0 x1 , f  x1 x0 ) ∧ (x0 = x1 → f x0 x1 = f  x1 x0 ) . Now we can apply the process described in the proof of Theorem 6.15. We obtain the following D-sentence:  ∀x0 ∀x1 ∃x2 ∃x3 =(x0 , x1 , x2 ) ∧ =(x1 , x0 , x3 ) ∧  φ(x0 , x1 , x2 , x3 ) ∧ (x0 = x1 → x2 = x3 ) .

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193

Exercise 6.21 Let M be a model, let φ define P ⊆ M n , and let ψ define Q ⊆ M n . We claim that φ ∧ ψ defines P ∩ Q and that φ ∨ ψ defines P ∪ Q. The proofs go as follows: (a1 , . . . , an ) ∈ P ∩ Q ⇐⇒ (a1 , . . . , an ) ∈ P and (a1 , . . . , an ) ∈ Q ⇐⇒ (M, a1 , . . . , an ) |= φ and (M, a1 , . . . , an ) |= ψ ⇐⇒ (M, a1 , . . . , an ) |= φ ∧ ψ; (a1 , . . . , an ) ∈ P ∪ Q ⇐⇒ (a1 , . . . , an ) ∈ P or (a1 , . . . , an ) ∈ Q ⇐⇒ (M, a1 , . . . , an ) |= φ or (M, a1 , . . . , an ) |= ψ ⇐⇒ (M, a1 , . . . , an ) |= φ ∨ ψ. The final equivalence follows from the truth definition of D-logic, remembering that M |= φ means M |={∅} φ and that {∅} = ∅ ∪ {∅}. To see why P \ Q is in general not D-definable even when P and Q are, first assume the contrary. Let Mω be the standard model of arithmetic. By Theorem 6.24 there is a sentence τ (c) of D such that for all sentences φ of D we have Mω |= φ if and only if Mω |= τ (φ). Also, let θ(c) be a sentence of D for which Mω |= θ(k) if and only if k is the G¨odel number of some sentence of D. The sentence θ(c) defines the set P ⊂ ω of G¨odel numbers of sentences of D, and the sentence τ (c) defines a set Q ⊂ ω that contains the G¨odel numbers of true sentences of D. By our assumption, the set P \ Q of G¨odel numbers of sentences of D that are not true is definable by a sentence π(c) of D. Therefore for all sentences φ of D we have the following: Mω |= φ if and only if Mω |= π(φ). By Theorem 6.19, π has a fixed point which is some sentence λ of D such that Mω |= λ if and only if Mω |= π(λ). Now we have that Mω |= λ if and only if Mω |= π(λ) if and only if Mω |= λ. This is a contradiction, completing the proof.

Exercise 6.22 Let ψ be a D-sentence in language L ∪ {R}. Assume we have models M and N with M |= ψ, N |= ψ, ML = N L, and R M = R N . Also assume that we have a D-sentence φ in language L ∪ {c1 , . . . , cn } that defines R in every model of ψ. We can assume w.l.o.g. that there is some (a1 , . . . , an ) ∈ R M \ R N . Then

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we have (a1 , . . . , an ) ∈ R M ⇒ (M, a1 , . . . , an ) |= φ ⇒ (ML , a1 , . . . , an ) |= φ ⇒ (N L , a1 , . . . , an ) |= φ ⇒ (N , a1 , . . . , an ) |= φ ⇒ (a1 , . . . , an ) ∈ R N , which is a contradiction, proving the claim.

Exercise 6.23 Let ψ be a D-sentence in the language L ∪ {R}. Assume that for all models M and N , if M |= ψ, N |= ψ and ML = N L, then R M = R N . Let ψ  be obtained from ψ by replacing occurrences of R by R  , where R  is a new relation symbol. From the assumption we get that when a model (M, A, B) satisfies the sentence ψ ∧ ψ  of the language L ∪ {R, R  }, then A = B. Therefore the D-sentences ψ ∧ Rc1 . . . cn and ψ  ∧ ¬R  c1 . . . cn have no models in common. We can then use the Separation Theorem and obtain a sentence φ in the language L ∪ {c1 , . . . , cn } such that every model of ψ ∧ Rc1 . . . cn is a model of φ, and that φ and ψ  ∧ ¬R  c1 . . . cn have no models in common. Now we have, for an arbitrary model M |= ψ in the language L ∪ {R}, the following implication: (a1 , . . . , an ) ∈ R M ⇒ (M, a1 , . . . , an ) |= ψ ∧ Rc1 . . . cn ⇒ (M, a1 , . . . , an ) |= φ. Also note that (M, A) |= ψ ⇐⇒ (M, A, A) |= ψ ∧ ψ  . We obtain the following implication for all models M |= ψ ∧ ψ  : (M, a1 , . . . , an ) |= φ ⇒ (M, a1 , . . . , an ) |= ¬R  c1 . . . cn ⇒ (a1 , . . . , an ) ∈ R M ⇒ (a1 , . . . , an ) ∈ R M .

Exercise 6.26 Denote by S the sentence “It is not true that S is true.” Assuming that truth is two-valued (which is the case for example in first order logic), S is either true or false. If S is true, then the contents of S state that S is false, which is a contradiction. On the other hand, if S is false, then the contents of S say that S is true, another contradiction. There are no other possibilities, so we have derived the required contradiction.

Exercise 6.27 If it is raining in both Warsaw and Vienna, then sentence (3) is not paradoxical because it can be false (or true) and still be consistent with sentences (1) and (2). Similarly, when it is not raining in Warsaw nor in Vienna, we can make (3) consistent by making it false. However, if it is raining in only one of Warsaw and Vienna, we cannot have (3) true or false and still have it consistent. Namely, if (3) is false, then exactly one of (1)–(3) is true, so (3) is actually true, a contradiction. Similarly, if (3) is true, then two of (1)–(3) are true, so (3) does not state a true condition, another contradiction.

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Exercise 6.28

  As in the proof of Theorem 6.19, let θ (x0 ) be ∃x1 φ(x1 ) ∧ σ (x0 , x1 , x0 ) , let k = θ(x0 ), and let ψ be θ (k). Then we have the following equivalence: Mω |= φ(ψ) ⇐⇒ Mω |= φ(θ (k)) ⇐⇒ there is a ∈ M with Mω |= φ(a) and (k, a, k) ∈ Sub   ⇐⇒ Mω |= ∃x1 φ(x1 ) ∧ σ (k, x1 , k) ⇐⇒ Mω |= θ (k) ⇐⇒ Mω |= ψ.

Exercise 6.31 Let T be a theory in the language L {+,×} ∪ {c0 , c1 , c2 , c3 }, where all the ci are new constant symbols, so that T consists of Peano’s axioms, an axiomatization of our auxiliary tools such as the relations SAT, POS-ID, NEG-ID, TRUE-ID, etc., and the sentences SATc0 c1 ∧ POS-IDc1 c2 c3 ∧ TRUE-IDc0 c2 c3 , and ¬(n = ck ) for all natural numbers n and all k = 0, 1, 2, 3. Then T expresses that the defined truth predicate will also say something about the truth of some non-standard G¨odel numbers. We can satisfy all finite subsets of T by a standard model (Mω , SatN ). By the Compactness Theorem of D we get that T has a model (M, S). Because of how we built T it now holds that even though S agrees with the standard truth predicate SatN about standard G¨odel numbers, there is also a pair of non-standard elements, namely (c0M , c1M ) ∈ S. That is why S = SatN , even though (M, S) |= θ L and (Mω , SatN ) |= θ L .

Exercise 6.33 We have the following inference for an arbitrary D-sentence φ to prove the first part of the exercise. Note that φ ↔ ¬φ is shorthand for (¬φ ∨ ¬φ) ∧ (¬¬φ ∨ φ): 

⇐⇒ ⇐⇒

⇐⇒ ⇐⇒

 (¬φ ∨ ¬φ) ∧ (¬¬φ ∨ φ), {∅}, 1 ∈ T     ¬φ ∨ ¬φ, {∅}, 1 ∈ T and ¬¬φ ∨ φ, {∅}, 1 ∈ T

     ¬φ, {∅}, 1 ∈ T or ¬φ, {∅}, 1 ∈ T and

     ¬¬φ, {∅}, 1 ∈ T or φ, {∅}, 1 ∈ T     ¬φ, {∅}, 1 ∈ T and φ, {∅}, 1 ∈ T     φ, {∅}, 0 ∈ T and φ, {∅}, 1 ∈ T

⇐⇒ {∅} = ∅. The inference ends in a contradiction. The reason why we can have a D-sentence λ that states “λ is not true” is that D is a three-valued logic, and that gives us more freedom. We avoid contradiction because λ is neither true nor false. Note that in D-logic, saying “φ is false” is different from saying

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Table A.7. I

Qc1

II

Rule

∃x0 P x0 Pc0 ∃x0 ¬Qx0 ¬Qc1

(1) (6) (1) (6) (4) (1) (4), (6), (4) (5) (4), (6), (4) (5)

∀x0 (P x0 ∨ Qx0 ) Pc0 ∨ Qc0 Pc0 Pc1 ∨ Qc1 Pc1 .. .

“φ is not true.” The latter expression is weaker than the former. In other words, when φ is false it is also not true, but there are cases when φ is not true but still not false.

Exercise 6.36 Assume there was some D-sentence τ  (c) such that for all D-sentences φ we had Mω |= φ iff Mω |= τ  (φ). Then by applying the D-alternative of G¨odel’s Fixed Point Theorem we get a D-sentence ψ such that Mω |= ψ iff Mω |= τ  (ψ). This is a contradiction with the first assumption.

Exercise 6.38 (i) Player II can always win by playing as shown in Table A.7. Note that even though we do not have an explicit rule for ∀xn φ, we can play it by rules (4) (for negation), (6) (for existential quantification), and (4) (for negation again) from Definition 6.25 because ∀xn φ is shorthand for ¬∃xn ¬φ. Player II can always choose Pcn from a disjunction Pcn ∨ Qcn . (ii) Player II can always win by playing as shown in Table A.8. All atomic sentences Rtt  in the game appear on the side of player II, so no contradiction can occur.

Exercise 6.39 (i) Player I can always win by playing as shown in Table A.9. Note that φ → ψ is shorthand for ¬φ ∨ ψ. At position (¬Pc0 ∨ Qc0 , II), player II could have chosen ¬Pc0 instead of Qc0 , but this would have led to an even easier victory for player I. (ii) Player I can always win by playing as shown in Table A.10.

Solutions to selected exercises, by Ville Nurmi

Table A.8. I ∃x∀y¬Rx y ∀y¬Rc0 y ¬Rc0 c1

II

Rule

¬∃x∀y¬Rx y

(1) (4) (6) (4), (6), (4) (4) (1) (4), (6), (4) (7) (3)

Rc0 c1 ∀x Rx f x Rc0 f c0 f c0 = c1 Rc0 c1 .. .

Table A.9. I

∃x0 Qx0 Qc0

II

Rule

∃x0 P x0 Pc0 ∀x0 (¬P x0 ∨ Qx0 ) ¬Pc0 ∨ Qc0 Qc0 ¬∃x0 Qx0

(1) (6) (1) (4), (6), (4) (5) (1) (4) (6)

Table A.10. I ∀x∃y Rx y ∃y Rc0 y

Rc0 c1

II

Rule

¬∀x∃y Rx y

(1) (4) (4), (6), (4) (1) (4), (6), (4) (7) (3) (6)

∀x Rx f x Rc0 f c0 f c0 = c1 Rc0 c1

197

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Exercise 6.40 Let be the finite set of sets S of pairs (φ, α) such that α ∈ {I, II}, φ is a sentence in vocabulary L  = L ∪ C, where C are countably many new constants, and L is a countable vocabulary, and for each S ∈ there is a model M S such that (a) the universe consists of constant interpretations, M S = {cMS : c ∈ C}; (b) if (φ, I) ∈ S then M S |= ¬φ; (c) if (φ, II) ∈ S then M S |= φ. We show that is a consistency property by going through conditions (i)–(ix) in the definition of a consistency property. (i) If S ∈ and t is a constant L  -term, then M S |= t = t (simply because any model would satisfy this sentence). Denote S  = S ∪ {(t = t, II)}. Now let us check if S  ∈ . All we need is to find a model M S  such that (a), (b), and (c) hold. Choose M S to be the candidate for M S  . Clearly (a) holds. To see (b), if (φ, I) ∈ S  , then (φ, I) ∈ S, so M S |= ¬φ. OK. To see (c), if (φ, II) ∈ S  , then either (φ, II) ∈ S or φ is t = t. In the first case we get M S |= φ. In the second case we again get M S |= φ, as we have already seen. This proves condition (i). (ii) Let S ∈ such that (t = t  , II) ∈ S, and let φ(t) be atomic. If (φ(t), II) ∈ S ∈ , then M S |= φ(t) and M S |= t = t  , so M S |= φ(t  ). Thus S ∪ {(φ(t  ), II)} ∈ . On the other hand, if (φ(t), I) ∈ S ∈ , then M S |= ¬φ(t) and M S |= t = t  , so M S |= ¬φ(t  ). Thus S ∪ {(φ(t  ), I)} ∈ . This proves condition (ii). (iii) If (¬φ, II) ∈ S ∈ , then M S |= ¬φ, so S ∪ {(φ, I)} ∈ . On the other hand, if (¬φ, I) ∈ S ∈ , then M S |= ¬¬φ, so M S |= φ, so S ∪ {(φ, II)} ∈ . This proves condition (iii). (iv) If (φ ∨ ψ, II) ∈ S ∈ , then M S |= φ ∨ ψ, so either M S |= φ or M S |= ψ. In the first case we get S ∪ {(φ, II)} ∈ , and in the latter case we get S ∪ {(ψ, II)} ∈ , which proves condition (iv). (v) If (φ ∨ ψ, I) ∈ S ∈ , then M S |= ¬(φ ∨ ψ), so both M S |= ¬φ and M S |= ¬ψ. Thus we get S ∪ {(φ, I)} ∈ and S ∪ {(ψ, I)} ∈ , which proves condition (v). (vi) If (∃xn φ(xn ), II) ∈ S ∈ , then M S |= ∃xn φ(xn ), so there is some a ∈ M S with M S |={xn →a} φ(xn ). By (a) there is some c ∈ C such that M S |= φ(c). Thus S ∪ {(φ(c), II)} ∈ , which proves condition (vi). (vii) If (∃xn φ(xn ), I) ∈ S ∈ , then M S |= ¬∃xn φ(xn ), so for all a ∈ M S we have M S |={xn →a} ¬φ(xn ). By (a), for all c ∈ C we have M S |= ¬φ(c). Thus S ∪ {(φ(c), I)} ∈ for all c ∈ C, which proves condition (vii). (viii) If S ∈ and t is a constant L  -term, then by (a) we have t MS = cMS for some c ∈ C. Thus M S |= t = c, so S ∪ {(t = c, II)} ∈ , which proves condition (viii). (ix) Assume there was some S ∈ such that (φ, II) ∈ S and (φ, I) ∈ S. Then M S |= φ and M S |= ¬φ, which is a contradiction. Thus the case of the assumption can never happen, which proves condition (ix). This concludes the proof that is a consistency property.

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199

Exercise 6.42 Let T be an L-theory. Assume that σ is a winning strategy for player II in MEG(T, L). We are to prove that there is a consistency property for T . The idea of the proof is that we create a set that imitates plays of MEG(T, L) when player II plays with her winning strategy σ against all the possible strategies of player I. Since all sets S ∈ must be finite in order to being a consistency property, we cannot take S to be all the positions in some play (might be infinite). Instead, we take only positions up to some finite number of moves. Precisely, let be the set such that S ∈ iff there is some strategy τ of player I in MEG(T, L) and natural number n < ω such that S is the set of positions of the play of MEG(T, L) when player I plays according to strategy τ and player II plays according to strategy σ . We prove that is a consistency property for theory T by going through conditions (i)–(ix) of the definition of a consistency property.

(i) Let S ∈ , obtained by some strategy τ for player I by playing up to some move n, and let t be a constant L  = L ∪ C-term. At this point in the game, player I can make an identity move on term t and the game moves to position (t = t, II). Thus S ∪ {(t = t, II)} ∈ . (ii) Let S ∈ , corresponding to some play at some phase, and let (φ(t), α) ∈ S, where φ(t) is atomic, and let (t = t  , II) ∈ S. Now player I can continue by making a substitution move, and the game continues from position (φ(t  ), α). Thus S ∪ {(φ(t  ), α)} ∈ . (iii) Let S ∈ , corresponding to some play at some phase. If (¬φ, II) ∈ S, then player I can make a negation move, and the game continues from position (φ, I). Thus S ∪ {(φ, I)} ∈ . On the other hand, if (¬φ, I) ∈ S, then player I can make a negation move such that the game continues from position (φ, II). Thus S ∪ {(φ, II)} ∈ . (iv) Let S ∈ , corresponding to some play at some phase, and let (φ ∨ ψ, II) ∈ S. Player I can make a disjunction move, and the game continues from position (φ, II) or (ψ, II), depending on player II’s choice. Thus either S ∪ {(φ, II)} ∈ or S ∪ {(ψ, II)} ∈ . (v) Let S ∈ , corresponding to some play at some phase, and let (φ ∨ ψ, I) ∈ S. Player I can make a disjunction move so that the game continues from position (φ, I). On the other hand, he can also choose the game to continue from position (ψ, I). Thus S ∪ {(φ, I)} ∈ and S ∪ {(ψ, I)} ∈ . (vi) Let S ∈ , corresponding to some play at some phase. Suppose (∃xn φ(xn ), II) ∈ S. Player I can make an existential move, and the game continues from position (φ(c), II) for some c ∈ C, depending on the choice of player II. Thus S ∪ {(φ(c), II)} ∈ for some c ∈ C. (vii) Let S ∈ , corresponding to some play at some phase. Suppose (∃xn φ(xn ), I) ∈ S. For each c ∈ C, player I is able to make an existential move so that the game continues from position (φ(c), I). Thus S ∪ {(φ(c), II)} ∈ for all c ∈ C. (viii) Let S ∈ , corresponding to some play at some phase, and let t be a constant L  term. Player I can make a constant move, and the game continues from position (t = c, II) for some c ∈ C, depending on the choice of player II.

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(ix) There cannot be any set S ∈ and atomic formula φ such that (φ, II) ∈ S and (φ, I) ∈ S, because S contains exactly the game positions up to some n moves in some play where player II has followed her winning strategy σ . Lastly, we check the requirement that the consistency property is a consistency property for theory T . Assume φ ∈ T and S ∈ . Then there is some strategy τ for player I in MEG(T, L) such that, after some n rounds of playing, the game has produced the set S of game positions. Certainly there is a strategy τ for player I such that on his next move he will perform a theory move and choose (φ, II) to become the new game position. Therefore S ∪ {(φ, II)} ∈ . This concludes the whole proof.

Exercise 6.44 Let H be a Hintikka set. Define, for c, c ∈ C, c ∼ c if (c = c , II) ∈ H . We show that ∼ is an equivalence relation. Firstly, for any c ∈ C, we have (c = c, II) ∈ H by condition (i) in the definition of a Hintikka set (Definition 6.30) because c is a constant L  -term. Thus ∼ is reflexive. Secondly, assume (c = c , II) ∈ H . By the above, we have (c = c, II) ∈ H . Now, by condition (ii) in Definition 6.30 we have (c = c, II) ∈ H . Thus ∼ is symmetric. Thirdly, assume (c = c , II) ∈ H and (c = c , II) ∈ H . Again, by condition (ii) we have (c = c , II) ∈ H . Thus ∼ is transitive.

Exercise 6.45 The proof follows conditions (i)–(ix) in the definition of a consistency property (Definition 6.28). In fact we have to generalize the definition slightly. Namely, in the original definition, all sentences that appear in are of some fixed vocabulary L ∪ C. In this exercise we do not have this situation. All sentences that occur in our are surely of the vocabulary L 1 ∪ L 2 ∪ C, but we avoid considering all such sentences. We have only L 1 ∪ C- and L 2 ∪ C-sentences. We must reflect this in the definition by slightly altering conditions (i) and (viii) which prompt for appearance of some sentence that contains an arbitrary term. In the following proof, we assume that the proper adjustments to the definition have been made. (i) Let S ∈ and let t be a constant L i ∪ C-term for some i ∈ {1, 2}. Denote S  = S ∪ {(t = t, II)}. If there were some L 1 ∩ L 2 -sentence θ that would separate T (S  )(L 1 ∪ C) and T (S  )(L 2 ∪ C), then θ would also separate T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C). Namely, if some model M has M |= T (S)(L 1 ∪ C), then M |= T (S  )(L 1 ∪ C), so M |= θ. Furthermore, if some M has M |= θ, then M |= T (S  )(L 2 ∪ C), so M |= T (S)(L 2 ∪ C). Therefore we have a contradiction to the fact that S ∈ , which says that T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C) are inseparable. Thus S  ∈ . (ii) Let (φ(t), α) ∈ S ∈ , let φ(t) be atomic, and let (t = t  , II) ∈ S. Denote S  = S ∪ {(φ(t  ), α)}. Assume φ(t) is an L 1 ∪ C-sentence or an L 2 ∪ C-sentence (it does not matter in this proof of condition (ii) which one it is). If there were some L 1 ∩ L 2 -sentence θ that would separate T (S  )(L 1 ∪ C) and T (S  )(L 2 ∪ C), then – with a proof similar to that of condition (i) – θ would also separate T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C), contrary to the fact that S ∈ . Thus S  ∈ .

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(iii) Let (¬φ, II) ∈ S ∈ . Assume that ¬φ is an L i ∪ C-sentence for some i ∈ {1, 2} (does not matter which). Denote S  = S ∪ {(φ, I)}. If it was the case that T (S  )(L 1 ∪ C) and T (S  )(L 2 ∪ C) were separable by some L 1 ∩ L 2 -sentence θ, then also T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C) would be separable by θ. The basic idea is that both (¬φ, II) and (φ, I) contribute the same sentence to T (S  ). More precisely, if we have a model M with M |= T (S)(L 1 ∪ C), then M |= T (S  )(L 1 ∪ C), so M |= θ . Furthermore, if some model M has M |= θ, then M |= T (S  )(L 2 ∪ C), so M |= T (S)(L 2 ∪ C). This is contrary to S ∈ . Thus we have S  ∈ . (iv) Let (φ ∨ ψ, II) ∈ S ∈ . Assume that φ ∨ ψ is an L 1 ∪ C-sentence.

(A.26)

Denote S  = S ∪ {(φ, II)} and S  = S ∪ {(ψ, II)}. If it were the case that we had L 1 ∩ L 2 -sentences θ  and θ  such that T (S  )(L 1 ∪ C) and T (S  )(L 2 ∪ C) are separable by θ  , and T (S  )(L 1 ∪ C) and T (S  )(L 2 ∪ C) are separable by θ  ,

(A.27)

then θ  ∨ θ  would separate T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C). Namely, assume that for some model M we have M |= T (S)(L 1 ∪ C). Then M |= φ ∨ ψ because of Eq. (A.26). Thus either M |= φ or M |= ψ. Then either M |= T (S  )(L 1 ∪ C) or M |= T (S  )(L 1 ∪ C), so by Eq. (A.27), M |= θ  or M |= θ  . In both cases we get M |= θ  ∨ θ  . Assume then that some model M has M |= θ  ∨ θ  . Then either M |= θ  or M |= θ  , so, by Eq. (A.27), either M |= T (S  )(L 2 ∪ C) or M |= T (S  )(L 2 ∪ C). Because of Eq. (A.26), we have T (S  )(L 2 ∪ C) = T (S)(L 2 ∪ C) and T (S  )(L 2 ∪ C) = T (S)(L 2 ∪ C). Thus we get, in both cases, M |= T (S)(L 2 ∪ C). This is a contradiction to S ∈ . Thus we have either S  ∈ or S  ∈ . The other case when φ ∨ ψ is an L 2 ∪ Csentence can be proved similarly. (v) Let (φ ∨ ψ, I) ∈ S ∈ . Assume Eq. (A.26). (The other case is proved similarly.) Denote S  = S ∪ {(φ, I)} and S  = S ∪ {(ψ, I)}. Assume there was some L 1 ∩ L 2 sentence θ such that either Eq. (A.28) or Eq. (A.29) holds: T (S  )(L 1 ∪ C) and T (S  )(L 2 ∪ C) are separable by θ,

(A.28)

T (S  )(L 1 ∪ C) and T (S  )(L 2 ∪ C) are separable by θ.

(A.29)

Then θ would separate T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C). For a proof, let us assume Eq. (A.28). The case for Eq. (A.29) is proved symmetrically. Now, if M |= T (S)(L 1 ∪ C), then M |= T (S  )(L 1 ∪ C) because of Eq. (A.26) and the fact that ¬(φ ∨ ψ) |= ¬φ. Thus, by Eq. (A.28), M |= θ. Furthermore, if some model M has M |= θ , then by Eq. (A.28) we have M |= T (S  )(L 2 ∪ C). Now, because of Eq. (A.26), T (S  )(L 2 ∪ C) = T (S)(L 2 ∪ C), so M |= T (S)(L 2 ∪ C). Therefore we get a contradiction with S ∈ . Thus we have S  ∈ and S  ∈ . (vi) Let (∃xn φ(xn ), II) ∈ S ∈ . Assume that ∃xn φ(xn ) is an L 1 ∪ C-sentence.

(A.30)

(The other case is proved similarly.) Let c ∈ C be a constant that does not occur in S (such constants exist because S is finite). Denote S  = S ∪ {(φ(c), II)}. If there

202

Solutions to selected exercises, by Ville Nurmi were some L 1 ∩ L 2 -sentence θ such that θ separates T (S  )(L 1 ∪ C) and T (S  )(L 2 ∪ C),

(A.31)

then θ would separate also T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C), contrary to S ∈ . More precisely, if some model M has M |= T (S)(L 1 ∪ C), then M |= ∃xn φ(xn ). Therefore there is some a ∈ M with (M, a) |= φ(c), so (M, a) |= T (S  )(L 1 ∪ C). By Eq. (A.31), (M, a) |= θ . Because θ does not mention c, we get M |= θ. Furthermore, if we have a model M such that M |= θ, then, by Eq. (A.31), M |= T (S  )(L 2 ∪ C). Because of Eq. (A.30), T (S  )(L 2 ∪ C) = T (S)(L 2 ∪ C), so we have M |= T (S)(L 2 ∪ C). This gives a contradiction with S ∈ . Thus we have S  ∈ . (vii) Let (∃xn φ(xn ), I) ∈ S ∈ . Assume Eq. (A.30). (The other case is proved similarly.) Denote, for each c ∈ C, Sc = S ∪ {(φ(c), I)}. If there were some L 1 ∩ L 2 sentence θ and some c ∈ C such that θ separates T (Sc )(L 1 ∪ C) and T (Sc )(L 2 ∪ C),

(A.32)

then θ would separate also T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C). Namely, if some model M has M |= T (S)(L 1 ∪ C), then M |= ¬∃xn φ(xn ). This gives us M |= ¬φ(c), so we have M |= T (Sc )(L 1 ∪ C). By Eq. (A.32) we get M |= θ. On the other hand, if some model M has M |= θ, then, by Eq. (A.32), we get M |= T (Sc )(L 2 ∪ C). Because of Eq. (A.30), we have T (Sc )(L 2 ∪ C) = T (S)(L 2 ∪ C), so M |= T (S)(L 2 ∪ C). This is contrary to S ∈ . Thus we have S  ∈ . (viii) Let S ∈ and assume that t is a constant L 1 ∪ C-term.

(A.33)

Let c ∈ C be a constant that does not occur in S (such constants exist because S is finite). Denote S  = S ∪ {(t = c, II)}. Now assume there is some L 1 ∩ L 2 sentence θ such that θ separates T (S  )(L 1 ∪ C) and T (S  )(L 2 ∪ C).

(A.34)

We prove that then θ separates also T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C), contrary to assumption S ∈ . Then Eq. (A.34) is false, and that gives us S  ∈ . So, assume that some model M has M |= T (S)(L 1 ∪ C). Denote a = t M ∈ M. Then (M, a) |= t = c, so (M, a) |= T (S  )(L 1 ∪ C). By Eq. (A.34), (M, a) |= θ. Because θ does not contain c, we get M |= θ . Assume then that some model M has M |= θ. Then, by Eq. (A.34), M |= T (S  )(L 2 ∪ C). Because of Eq. (A.33), T (S  )(L 2 ∪ C) = T (S)(L 2 ∪ C), so M |= T (S)(L 2 ∪ C). (ix) Let S ∈ and assume there is some L 1 ∪ C-sentence φ such that (φ, II) ∈ S and (φ, I) ∈ S. Let θ be the L 1 ∩ L 2 -sentence ¬∀x0 (x0 = x0 ). Now, if some model M has M |= T (S)(L 1 ∪ C), then M |= θ. This is so because φ is an L 1 ∪ C-sentence, so T (S)(L 1 ∪ C) contains both sentences φ and  φ. Thus T (S)(L 1 ∪ C) has no models. Similarly, because θ has no models, we get that θ and T (S)(L 2 ∪ C) have no models in common. This proves that θ separates T (S)(L 1 ∪ C) and T (S)(L 2 ∪ C), contrary to S ∈ . The case that φ is an L 2 ∪ C-sentence is similar. This concludes the whole proof.

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Table A.11. Team X for Exercise 6.54 x0

x1

x2

0 1 2

2 0 1

1 0 1

Table A.12. Teams X and Y for Exercise 6.57 X

Y

x0

x1

x2

x0

x1

x2

0 1 2

2 0 1

2 0 1

a b c d

c b a d

b c a d

Exercise 6.54 Fml03

The set contains infinitely many formulas, but only a finite number of them are non-equivalent. We confine ourselves to list only one formula from each equivalence class that X (see Table A.11) satisfies in M: 

¬(x0 = x1 )

¬(x0 = x2 )

=(x0 , x1 )

=(x0 , x2 )

=(x1 , x0 )

=(x1 , x2 ).

In addition to these, there are all the conjunctions of any subset of these listed formulas, such as ¬(x0 = x1 ) ∧ =(x0 , x1 ) ∧ =(x1 , x2 ). The team X also satisfies trivial consequences of these listed formulas. For example, because =(x0 , x2 ) is satisfied, then also =(x0 , x1 , x2 ) is satisfied because we have =(x0 , x2 ) ⇒ =(x0 , x1 , x2 ). Remember that formulas such as =(x0 , x0 ), =(x0 , x1 , x0 ), =(x0 , x0 , x0 , x0 ) and so on are satisfied by any team in any model, thus they are equivalent with . Also remember that formulas such as ¬ =(x2 , x0 ) are not satisfied by X in M. In fact, negations of determination formulas are always false: remember the truth definition! Fml03 also does not contain any disjunctions because their rank would be greater than 0.

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Table A.13. Teams X = X 0 ∪ X 1 and Y for Exercise 6.59 X

X0

X1

Y

x0

x1

x0

x1

x0

x1

x0

x1

0 1 2 3

2 3 3 2

1 2

3 3

0 3

2 2

a b c d

b c d a

Table A.14. Teams X , X (F/x2 ), and Y for Exercise 6.60 X

X (F/x2 )

Y

x0

x1

x0

x1

x2

x0

x1

0 1 2

2 1 1

0 1 2

2 1 1

0 0 0

a b c

b b a

Exercise 6.57 Player I won the game because, for example, M |= X x1 = x2 and N |=Y x1 = x2 . Another reason why player I won is that M |= X ¬(x0 = x1 ) and N |=Y ¬(x0 = x1 ). There may be other similar reasons too. See Table A.12.

Exercise 6.59 A good splitting move for player I is to present X as X 0 ∪ X 1 , as seen in Table A.13. Then player II must present Y as Y0 ∪ Y1 . After this player I necessarily wins, because M |= X 0 =(x1 ) and M |= X 1 =(x1 ), but whatever Y0 and Y1 are, for one of them we have N |=Yi =(x1 ). You can see this by checking all the relevant 16 different splits Y0 ⊂ Y , Y 1 = Y \ Y0 .

Exercise 6.60 A good supplementing move for player I is to consider the function F : X → M that maps all s ∈ X as F(s) = 0 (see Table A.14). When player I plays X (F/x2 ), he wins no matter which team Y (G/x2 ) player II plays. Namely, after this move we have M |= X (F/x2 ) P x2 but N |=Y (G/x2 ) P x2 . The latter comes simply from the fact that P N = ∅.

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Exercise 6.63 Let N = (N, +, ·, 0, 1, <), the standard model of arithmetic. Recall that there is a sentence N that is true in a model iff the model is not isomorphic to N . Now, if M is any uncountable model, it cannot be isomorphic to N . Thus M |= N , but N is isomorphic to itself, so N |= N . Thus we do not have N ≡D M.

Exercise 6.64 We can in principle show (R, N) D (Q, N) by giving, for each natural number n, a winning strategy for player II in EFn ((R, N), (Q, N)). However, winning strategies in the Ehrenfeucht–Fra¨ıss´e game for D are in general very difficult to describe. Luckily, there is an easier proof. First we have to generalize the L¨owenheim–Skolem Theorem of D (Theorem 6.5) so that it speaks about a whole theory instead of just a single sentence. L¨owenheim–Skolem Theorem of D Suppose T is a set of D-sentences such that T has an infinite model or arbitrarily large finite models. Then T has models of all infinite cardinalities, in particular T has a countable model and an uncountable model. Proof Enumerate the theory as T = {φk : k < ω}. For each k < ω, let τ1,φk be the 11 -translation of φk , τ1,φk = ∃S1k · · · ∃Snk ψk , where ψk is first order in the vocabulary L ∪ {S1k , . . . , Snk }. We can assume that all the relation symbols Sik are distinct. Denote L  = L ∪ {Sik : k < ω, i < n}. By the L¨owenheim–Skolem Theorem of first order logic, there is an L  -model M of the theory {ψk : k < ω} such that M is of cardinality κ. The reduction M = M L of M to the original vocabulary L is a model of T of cardinality κ.  With this formulation of the L¨owenheim–Skolem Theorem, we get that, letting T denote the D-theory of the model (R, N), T = {φ : (R, N) |= φ}, there is a countable model M of T . Recall that there is a D-sentence ∞ that is true in a model N iff N is infinite. With slight modifications we can also write down a sentence  that is true in a model N of language {P} iff both P N and N \ P N are infinite. This sentence  is then in the theory T . Thus M |= , and hereby M is isomorphic to (Q, N). Thus also (Q, N) |= T . This shows (R, N) D (Q, N).

Exercise 6.71 Let P and P  be dependence back-and-forth sets for M D M and M D M , respectively. To show M D M , define P  as follows: P  = {(X, Z ) : there is a team Y such that (X, Y ) ∈ P and (Y, Z ) ∈ P  }. We will now show that P  is a dependence back-and-forth set for M D M . Firstly, P  ⊆ Part(M, M ) because if (X, Z ) ∈ P  then we have some (X, Y ) ∈ P and (Y, Z ) ∈ P  , so there is n such that X ⊆ M n and Y ⊆ M n , and m such that Y ⊆ M m and Z ⊆ M m . This means m = n, so X ⊆ M n and Z ⊆ M n . Also, if M |= X φ for some atomic formula φ, then by assumptions, M |=Y φ, and further, M |= Z φ. To see that P  is non-empty, first show that (∅, ∅) ∈ P and (∅, ∅) ∈ P  . We know that P is non-empty, so there is some (X, Y ) ∈ P. Because X = X ∪ ∅, there is some split Y = Y1 ∪ Y2 such that (∅, Y2 ) ∈ P. Because M |=∅ ⊥, we have M |=Y2 ⊥. This

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means that Y2 = ∅. Thus we have (∅, ∅) ∈ P. Similarly for P  . Thus we get (∅, ∅) ∈ P  , which shows that P  is non-empty. Condition (i) Let (X, Z ) ∈ P  and let X = X 1 ∪ X 2 . By definition of P  , we then have some (X, Y ) ∈ P and (Y, Z ) ∈ P  . By properties of P there is some split Y = Y1 ∪ Y2 such that (X 1 , Y1 ) ∈ P and (X 2 , Y2 ) ∈ P. Again, by properties of P  there is some split Z = Z 1 ∪ Z 2 such that (Y1 , Z 1 ) ∈ P  and (Y2 , Z 2 ) ∈ P  . By definition of P  we get (X 1 , Z 1 ) ∈ P  and (X 2 , Z 2 ) ∈ P  . Condition (ii) Let (X, Z ) ∈ P  and let n be a natural number. By definition of P  we then have some (X, Y ) ∈ P and (Y, Z ) ∈ P  . By properties of P and P  we have (X (M/xn ), Y (M  /xn )) ∈ P, and further, (Y (M  /xn ), Z (M  /xn )) ∈ P  . By definition of P  we get (X (M/xn ), Z (M  /xn )) ∈ P. Condition (iii) Let (X, Z ) ∈ P  , let n be a natural number, and let F : M → X be some function. By definition of P  we then have some (X, Y ) ∈ P and (Y, Z ) ∈ P  . By properties of P there is some function G : M  → Y such that (X (F/xn ), Y (G/xn )) ∈ P. By properties of P  there is some function H : M  → Z such that (Y (G/xn ), Z (H/xn )) ∈ P  . Thus, by definition of P  we get (X (F/xn ), Z (H/xn )) ∈ P  . This completes the proof.

Chapter 7 Exercise 7.1 Here are the definitions as 0 -formulas. For brevity, we write ∀x ∈ yφ for ∀x(x ∈ y → φ), and we write ∃x ∈ yφ for ∃x(x ∈ y ∧ φ): x⊆y∪z

(i) (ii) (iii) (iv)

x x x x

y⊆x

z⊆x



 

 

  = y∪z ∀v ∈ x(v ∈ y ∨ v ∈ z) ∧ ∀v ∈ y(v ∈ x) ∧ ∀v ∈ z(v ∈ x), = {y, z} ∀v ∈ x(v = y ∨ v = z) ∧ y ∈ x ∧ z ∈ x, is transitive ∀y ∈ x∀z ∈ y(z ∈ x), is an ordinal “x is transitive” ∧ ∀y ∈ x(“y is transitive”), x is a relation between y and z

(v) x : y → z



  ∀v ∈ x∃u ∈ y∃w ∈ z(v = (u, w)) ∧ every element in y has an image



  ∀u ∈ y∃w ∈ z∃v ∈ x(v = (u, w)) ∧ images are unique



   ∀u ∈ y∀r ∈ z∀s ∈ z (∃v ∈ x(v = (u,  r ))∧ ∃v ∈ x(v = (u, s))) → r = s . Definition (iv) uses definition (iii). For definition (v) we also need to define “v = (u, w)” as a 0 -formula. A suitable formula is z = {u, {u, w}}, which in turn uses definition (ii) above.

Exercise 7.2 We prove the claim by induction on φ.

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If φ is an atomic formula, it is of the form xi = x j or xi ∈ x j . In both cases the claim holds. For example, (M, ∈) |= ai ∈ a j iff ai ∈ a j , simply because the model (M, ∈) interprets the binary relation symbol as the usual “belongs to” relation. The cases for negation, disjunction and existential quantifier go as follows: (M, ∈) |= ¬φ(a1 , . . . , an ) ⇐⇒ (M, ∈) |= φ(a1 , . . . , an ) ⇐⇒ not φ(a1 , . . . , an ) ⇐⇒ ¬φ(a1 , . . . , an ); (M, ∈) |= (φ ∨ ψ)(a1 , . . . , an ) ⇐⇒ (M, ∈) |= φ(a1 , . . . , an ) or (M, ∈) |= ψ(a1 , . . . , an ) ⇐⇒ φ(a1 , . . . , an ) or ψ(a1 , . . . , an ) ⇐⇒ (φ ∨ ψ)(a1 , . . . , an ); (M, ∈) |= ∃x(x ∈ ai ∧ φ)(x, a1 , . . . , an ) ⇐⇒ there is a ∈ M such that (M, ∈) |= a ∈ ai ∧ φ(a, a1 , . . . , an ) ⇐⇒ there is a ∈ M : (M, ∈) |= a ∈ ai and (M, ∈) |= φ(a, a1 , . . . , an ) ⇐⇒ there is a ∈ M such that a ∈ ai and φ(a, a1 , . . . , an ) ⇒ there is a such that a ∈ ai and φ(a, a1 , . . . , an )

(A.35) (A.36)

⇐⇒ there is a such that a ∈ ai ∧ φ(a1 , . . . , an ) ⇐⇒ ∃x(x ∈ ai ∧ φ)(x, a1 , . . . , an ). To achieve the implication Eq. (A.36) ⇒ Eq. (A.35), assume Eq. (A.36). Because ai ∈ M and M is transitive, also a ∈ M. Thus we get Eq. (A.35).

Exercise 7.3 Let (M, E) be a well-founded model that satisfies the Axiom of Extensionality, ∀x0 ∀x1 (∀x2 (x2 ∈ x0 ↔ x2 ∈ x1 ) → x0 = x1 ). First of all, just as we have defined the rank of an element, rk(x) = sup{rk(y) + 1 : y ∈ x}, we will now define the E-rank of an element x ∈ M by rk E (x) = sup{rk E (y) + 1 : y E x}. Because (M, E) is well-founded, we can define a mapping π by the equation π(x) = {π(y) : y E x} by induction on the E-rank of elements x ∈ M. Namely, if rk E (x) = 0, then there are no y ∈ M with y E x, so we can set π (x) = ∅. Let x ∈ M and assume that π has been defined for elements with E-rank lower than rk E (x), in particular for y ∈ M with y E x. Then the equation π(x) = {π (y) : y E x} is well defined, so we can use it for defining π(x). Denote by N the image of M under π , N = π  M. We claim that N is a transitive set. To prove it, let x ∈ N and y ∈ x. Then y = π(z) for some z ∈ M, so y ∈ N , by definition of π . Next we claim that π is one-to-one. As an assumption towards contradiction, assume π(x) = π(y) for some x, y ∈ M with x = y. In fact, we can and will pick π(x) and π(y) so that they are also minimal in respect of their rank, rk(π (x)). Because (M, E) is extensional, we get from x = y that there is u ∈ M with u ∈ x and u ∈ y (or u ∈ x and

208

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u ∈ y, but that is a symmetric case). Of course π(u) ∈ π(x), and because π(x) = π (y) we also get that π(u) ∈ π(y). But by definition of π there is some v ∈ y such that π (v) = π (u). Now we have u = v and π(u) = π(v), and because π(u) ∈ π (x) we also have rk(π (u)) < rk(π(x)). This is contrary to our choice of π(x) to be minimal in respect of rank. Thus π is one-to-one. As N is merely the image of M under π , π : M → N is even a bijection. To see that π is an isomorphism, we need to show x E y ⇐⇒ π(x) ∈ π(y). The implication to the right is clear from the definition of π. For the other direction, assume π(x) ∈ π (y). By definition of π , there is some z ∈ M such that z E y and π(z) = π(x). Because π is one-to-one, we get that z = x. Thus x E y. Our final claim is that if E is actually ∈, the usual “belongs to” relation, then for any transitive set A ⊆ M we have π(x) = x for all x ∈ A. We prove the claim by induction on the rank of elements. Note: because E is ∈, we can write things like y ∈ x instead of y E x for elements x, y ∈ M. In general, for all x ∈ M we have y E x iff y ∈ x and y ∈ M iff y ∈ x ∩ M. In particular, the definition of π becomes π(x) = {π(y) : y ∈ x ∩ M}. Now, let x ∈ A. Because A is transitive, we get x ⊆ A, and thus x = x ∩ A = x ∩ M. If rk(x) = 0, then x = ∅, and by definition of π, π(x) = ∅. Otherwise, we assume that the claim holds for elements y ∈ A with rk(y) < rk(x), in particular all y ∈ A with y ∈ x, that is, all y ∈ x ∩ A = x ∩ M. Then we have the following equation: π(x) = {π(y) : y ∈ x ∩ M} = {y : y ∈ x ∩ M} = x ∩ M = x. This completes the proof.

Exercise 7.4 Our goal is to find a 2 -definition φ(x) for the property of x being equal to Vκ , where κ = κ . Let us first express φ(x) as a formula and then explain it in words: φ := φ1 ∧ φ2 ∧ ∀zφ3 ∧ ∃k∃V ∃ f (φ4 ∧ φ5 ); φ1 := Trans(x) ∧ ∃y ∈ x(y = ω); φ2 := ∀u ∈ x∃y ∈ x(y = Px (u)); φ3 := ∀y ∈ x(z ⊆ y → z ∈ x);     φ4 := ∀y ∈ x Ord(y) → y ∈ k ∧ ∀y ∈ k y ∈ x ∧ Ord(y) ∧ Bij( f, x, k);   φ5 := Fun(V, k, x) ∧ ∀z ∈ x∃y ∈ k z ∈ V (y) ∧   ∀y ∈ k y = ∅ → V (y) = ∅ ∧   ∀y ∈ k∀z ∈ k z = y + → V (z) = Px (V (y)) ∧   ∀y ∈ k Limit(y) → ∀z ∈ y(V (z) ⊆ V (y)) ∧  ∀u ∈ V (y)∃z ∈ y(u ∈ V (z)) . The formulas Trans(x), Ord(x), and Fun(x, y, z) are 0 -definitions for the properties “x is transitive,” “x is an ordinal,” and “x is a function y → z,” respectively, as in Exercise 7.1. The other relevant 0 -formulas and informal expressions are as

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209

follows: x =∅

∀y ∈ x(y = y);

x⊆y

∀u ∈ x(u ∈ y);

y = Px (z)

∀u ∈ x(u ∈ y ↔ u ⊆ z);

x = y+

Bij(x, y, z)

∀u ∈ x(u ∈ y ∨ u = y) ∧ y ⊆ x ∧ y ∈ x;   ¬(x = ∅) ∧ Ord(x) ∧ ∀y ∈ x Ord(y) → ∃z ∈ x(z = y + ) ;   Limit(x) ∧ ∀y ∈ x Limit(y) → y = ∅ ;   Fun(x, y, z) ∧ ∀u ∈ y∀v ∈ y x(u) = x(v) → u = v

ψ( f (y))

∧ ∀u ∈ z∃v ∈ y(x(v) = u);   ∃ p ∈ f ∃v ∈ p p = (y, v) ∧ ψ(v) .

Limit(x) x =ω

On the last line, ψ(x) is a 0 -formula, and f is a variable symbol for which Fun( f, r, s) ∧ y ∈ r holds for some variable symbols r and s. We see that φ1 , . . ., φ5 are 0 -formulas. Therefore φ is equivalent to a 2 -formula. Note that φ itself is not, strictly speaking, a 2 -formula because the unrestricted quantification does not appear only at the very beginning of the formula. However, this can be cured simply by pulling out the unrestricted quantifiers from the conjuncts. The subformula φ1 (x) says that x is transitive and contains ω; φ2 (x) says that x satisfies the power set axiom; ∀zφ3 (x) says that x contains all subsets of its elements. We will need this to ensure that the power sets that x contains are really power sets in the usual sense. Further, φ4 (x) says that k is the set of ordinals in x, and that f is a bijection between the ordinals of x and x itself; φ5 (x) says that V : k → x is a restriction of the function α → Vα for ordinals α, and that x is covered by the values of the function V . We see  that,when κ = κ , Vκ satisfies φ(x). For example, Vκ satisfies the formula ∃k∃V ∃ f φ4 (x) because the ordinals in Vκ are exactly all α ∈ κ, and, on the other hand, it holds in general (because κ is an ordinal) that |Vω+κ | = κ . Because ω = ω , we have ω ∈ κ, so ω + κ = κ. Now we have |Vκ | = |Vω+κ | = κ = κ, so there is a bijection between Vκ and κ, the ordinals in Vκ . To see that no other sets satisfy φ(x), assume that φ(M) holds for some set M. From φ4 (M) we get that k is the set of ordinals in M. From φ1 (M) we get that M is transitive. Because also the class ON of ordinals is transitive, the set k = M ∩ ON is transitive, and thus k is an ordinal itself. From φ1 (M) we get ω < k. We also get from φ4 (M) that |M| = |k|. We claim that for all ordinals α ∈ k we have V (α) = Vα , and prove it by induction on α. From φ5 (M) we get that V (∅) = ∅ = V∅ . For a successor ordinal α + ∈ k, assuming that V (α) = Vα , we get V (α + ) = {z ∈ M : z ⊆ Vα }. From ∀zφ3 (M) we get {z : z ⊆ Vα } ⊆ {z ∈ M : z ⊆ Vα }. Thus we get V (α + ) = {z ∈ M : z ⊆ Vα } = {z : z ⊆ Vα } = P(Vα ) = Vα+ . For a limit  ordinal γ ∈ k, assuming  that V (α) = Vα for all α < γ , we get from φ5 (M) that V (γ ) = {V (α) : α < γ } = {Vα : α < γ } = Vγ . This completes the induction. We show that k is not a successor ordinal. Let α ∈ k. Then Vα ∈ M. By φ2 (M) and ∀zφ3 (M), Vα+ = P(Vα ) = P M (Vα ) ∈ M. Because α + ⊆ Vα+ , we get by ∀zφ3 (M) that α + ∈ M, whence α + ∈ k. Hereby k must be a limit ordinal.

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 Now we have Vk = {Vα : α < k}, and for each α < k we have Vα ⊆ M by the facts that  Vα ∈ M and M istransitive. Therefore Vk ⊆ M. On the other hand, by φ5 (M), M ⊆ {V (α) : α ∈ k} = {Vα : α < k} = Vk . Thus M = Vk . This, combined with the previous results, yields |Vk | = |k|. This in addition to the fact that ω < k shows that k is an uncountable ordinal. Therefore ω + k = k, whence k = |Vω+k | = |Vk | = |k|. Thus k is a cardinal and k = k . We have shown that M = Vκ for some κ = κ .

Exercise 7.5 To solve the exercise it suffices to prove the more general claim that, for all D-formulas φ, all limit ordinals α, all M ∈ Vα , and all X ∈ Vα , we have Satφ (M, X )

iff

Vα |= Satφ (M, X ).

(A.37)

Here, Satφ (M, X ) is the first order conjunction “M is a model” ∧ “X is a team for M” ∧ “Fr(φ) ⊆ dom(X )” ∧ “M |= X φ.” To be very very precise, we would have to write out Satφ (M, X ) explicitly as a first order formula and prove Eq. (A.37), like we did in Exercise 7.2 for all 0 -formulas. The difference is that this time we will face some formulas that are not 0 , but luckily we also have a stronger assumption, namely that M, X ∈ Vα and α is a limit ordinal. However, we shall avoid this tedious task for the most part, and show just an outline and a few excerpts from the proof. We can express “M is a model (in the language of φ)” as a 0 -formula by writing something like “M is a tuple (a1 , . . . , an ), where a2 , . . . , an are constants, relations and functions of the correct arities in respect of the language of φ.” From Exercise 7.2, we get that “M is a model” iff

Vα |= “M is a model.”

Borrowing the formula Fun(x, y, z) from   Exercise 7.4, we can express “X is a team for M” by writing ∃d∀s ∈ X Fun(s, d, M) . This is not a 0 -formula, but d corresponds to dom(X ) and we know that rk(dom(X )) ≤ rk(X ). Thus from X ∈ Vα we get that if we find a suitable value for d from the set theoretical universe, we find this value in Vα . This yields “X is a team for M”

iff

Vα |= “X is a team for M.”

We can write “Fr(φ) ⊆ dom(X )” in a similar way. Note that φ is fixed relative to the formula Satφ (M, X ), and so is the finite set Fr(φ). Therefore it is very simple to write a formula that defines Fr(φ). The most demanding part is to define, for each D-formula φ, a first order formula Modφ (M, X ) that expresses “M |= X φ.” We can define Modφ (M, X ) inductively on φ. The definitions mimic the truth definition of D. A few examples are as follows:  M M Mod=(t1 ,...,tn ) := ∀s ∈ X ∀s  ∈ X (t1M s = t1M s   ∧ · · · ∧ tn−1 s = tn−1 s  )  → tnM s = tnM s   ;   Modφ∨ψ := ∃Y ∃Z X = Y ∪ Z ∧ Modφ (M, Y ) ∧ Modψ (M, Z ) ;   Mod∃xφ := ∃F∃Y Fun(F, X, M) ∧ Y = X (F/x x ) ∧ Modφ (M, Y ) .

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211

We now proceed to a proof by induction on φ that Modφ (M, X ) ⇐⇒ Vα |= Modφ (M, X ).

(A.38)

We shall only present the proof for the three cases mentioned above. If φ is =(t1 , . . . , tn ), then Modφ (M, X ) is a 0 -formula as soon as we write out a 0 -formula expressing “x = t M s” for a term t. It is possible but takes some space and is therefore omitted here. Now Exercise 7.2 applies, and we get Eq. (A.38). If φ is ψ ∨ θ, then any suitable values for the variables Y and Z in the formula Modφ (M, X ) are bound to be in Vα because X ∈ Vα , rk(Y ) ≤ rk(X ) and rk(Z ) ≤ rk(X ). Thus we get Eq. (A.38). Let φ be ∃xψ. We will use the fact that if rk(dom( f )) ≤ β and rk(rng( f )) ≤ β then rk( f ) < β + ω. Because M, X ∈ Vα and α is a limit, there is some β < α such that rk(M) ≤ β and rk(X ) ≤ β. Therefore any suitable value for the variable F in Modφ (M, X ) has rk(F) < β + ω ≤ α. Thus F ∈ Vα . We need also Y = X (F/x) ∈ Vα . Because dom(s) is a finite collection of variable symbols, we can assume that all s ∈ Y have rk(dom(s)) = m < ω, where m does not depend on s. We also know that rk(rng(s)) ≤ β, as rng(s) ⊆ M. Thus rk(s) ≤ β + n, where n < ω does not depend on s. Combining these facts, we get that rk(Y ) = sup{rk(s) + 1 : s ∈ Y } ≤ β + n + 1 < α. Thus any suitable value for the variable Y in Modφ (M, X ) has Y ∈ Vα . Therefore we get Eq. (A.38).

Exercise 7.6 Our goal is to write down a D-sentence ψ such that, for all models M, M |= ψ iff M is not isomorphic to a model (Vκ , ∈, K , V, f ), where κ = κ , K = {α : α < κ}, V : K → Vκ maps α → Vα , and f : M → K is a bijection. Note that this property is in close correspondence with the one in Exercise 7.4, where we wrote down a 2 -formula φ(x) that is satisfied only by Vκ , where κ = κ . Let us examine deeper the analogy between Exercises 7.4 and 7.6. The variable symbol x in Exercise 7.4 corresponds to the arbitrary model M in Exercise 7.6. Bounded quantification in the formula φ(x) corresponds to picking an element in the model M. This can be done by first order quantification in the D-sentence ψ. Unbounded quantification in φ(x) corresponds to picking some object outside the universe M. This can, in certain cases, be seen as second order quantification, i.e. the use of dependency statements =(t1 , . . . , tn ) in ψ. However, we can only express second order existential quantification in D, and there are both existential and universal unbounded quantifiers in φ(x). More precisely, we know how to express the existence of functions f : M → M and subsets A ⊆ M in D. Unbounded universal quantifiers in φ(x) can be expressed by an extension of the language (i.e. vocabulary) of M. Because M is arbitrary, the interpretations of any extra non-logical symbols are also arbitrary, so this will correspond to universal quantification of certain kind of elements outside the universe M, and those elements are functions f : M → M and subsets A ⊆ M. Luckily, this will be enough. This said, we take as a starting point the formula φ(x) from the solution of Exercise 7.4. Remember that ψ is to express practically ¬φ(x). Let the language of ψ be {E, K , V, f }. The binary relation symbol E shall denote the “belongs to” relation of M, and the other symbols are as in Exercise 7.4; K is a unary relation symbol, and

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Solutions to selected exercises, by Ville Nurmi

V and f are unary function symbols. Let ψ be as follows (further explanations can be found later): ψ := ¬ψ1 ∨ ¬ψ2 ∨ ψ3 ∨ ¬ψ4 ∨ ¬ψ5 ∨ ¬ψ6 ∨ wf ; ψ1 := ∃x(x = ω); ψ2 := ∀x∃y(y = P M (x));  ψ3 := ∃x∃w∀u∃v ∀z((u = z → v = w) → z E x)   ∧ ∀y∃z =(y, z) ∧ ¬((u = z → v = w) ↔ z E y) ; ψ4 := ∀x(K y ↔ Ord(y)) ∧ Bij( f, K );   ψ5 := ∀x∃y K y ∧ x E(V y) ∧ ∀x(x = ∅ → V x = ∅)   ∧ ∀x∀y (K x ∧ y = x + ) → V y = P M (V x)    ∧ ∀x Limit(x) → ∀y(y E x → V y ⊆ V x) ∧ ∀z∃y(z ∈ V x → z ∈ V y) ;   ψ6 := ∀x∀y ∀z(z E x ↔ z E y) → x = y . Note that wf is the formula that holds in a model M = (M, E) iff the binary relation E M is not well-founded. The abbreviations used above are written in D as follows: x =∅

∀y¬(y E x);

x⊆y

∀u(u E x → u E y);

y = P M (x)

∀u(u E y ↔ u ⊆ x);

x = y+

∀u(u E x ↔ (u E y ∨ u = y));

Trans(x)

∀y∀z(z E y → z E x);

Ord(x) x =ω

Trans(x) ∧ ∀y(y E x → Trans(y));   ¬(x = ∅) ∧ Ord(x) ∧ ∀y (y E x ∧ Limit(y)) → ∃z(z E x ∧ z = y + ) ;   Limit(x) ∧ ∀y (y E x ∧ Limit(y)) → y = ∅ ;

Bij( f, P)

∀u(P f u) ∧ ∀u∀v( f u = f v → u = v) ∧ ∀u∃v(Pu → f v = u).

Limit(x)

Note that Bij( f, P) expresses that the unary function f M is a bijection M → P M . The subformula ψ2 is the power set axiom; ψ3 expresses that “M misses a subset,” i.e. there is an element x ∈ M and a set A ⊆ M of elements z that belong to x in the sense of the model M, but there is no element y ∈ M such that the elements of A would be exactly the elements of y in the sense of the model M; ψ4 expresses that K M is the set of ordinals in M in the sense of M and f M is a bijection M → K M ; ψ5 expresses that M is covered by the values of the function V M and that V M is a restriction of the function α → Vα for ordinals α in the sense of M; ψ6 is the axiom of extensionality. We see that (Vκ , ∈, K , V, f ) |= ψ when κ = κ , K = {α : α < κ}, V : K → Vκ maps α → Vα , and f : M → K is a bijection. Let us check, for example, how (Vκ , ∈ , K , V, f ) |= ψ3 . Let x, w ∈ Vκ , and let F : Vκ → Vκ choose for each value u ∈ Vκ some value v = F(u) such that u ∈ x whenever F(u) = w. Let y = {z ∈ Vκ : F(z) = w}. We have rk(z) < rk(x) < κ for all z ∈ y, whence rk(y) < κ, so y ∈ Vκ . Now, if z ∈ Vκ , we have F(z) = w iff z ∈ y. This shows that ψ3 fails. Assume then that M = (M, E, K , V, f ) is a model for which M |= ψ. We want to show that M is isomorphic to some model (Vκ , ∈, K  , V  , f  ) with κ = κ . Because formulas ψi for i = 1, 2, 4, 5, 6 are first order, we have N |= ¬ψi iff N |= ψi for these

Solutions to selected exercises, by Ville Nurmi

213

i. From M |= ψ6 we get that M is extensional, and from M |= wf we get that M is well-founded. By Mostowski’s Collapsing Lemma, M is isomorphic to a transitive model N = (N , ∈, K  , V  , f  ). From N |= ψ4 we get that K  is the set of ordinals in N , and thus K  is an ordinal itself. From N |= ψ1 we get that ω < K  . We also get from N |= ψ4 that |N | = |K  |. We claim that for all ordinals α ∈ K  we have V  (α) = Vα , and prove it by induction on α. From N |= ψ5 we get that V  (∅) = ∅ = V∅ . For a successor ordinal α + ∈ K  , assuming that V  (α) = Vα , we get V  (α + ) = {z ∈ N : z ⊂ Vα }. From N |= ψ3 we get that whenever z ⊂ Vα , then z ∈ N . Thus we get V  (α + ) = {z ∈ N : z ⊂ Vα } = {z : z ⊂  Vα } = P(Vα ) = Vα+ . For a limit ordinal K  , assuming that γ ∈  V (α) = Vα for all α <   γ , we get from N |= ψ5 that V (γ ) = {V (α) : α < γ } = {Vα : α < γ } = Vγ . This completes the induction. We show that K  is not a successor ordinal. Let α ∈ K  . Then Vα ∈ N . By N |= ψ2 and N |= ψ3 , Vα+ = P(Vα ) ∈ N . Because α + ⊆ Vα+ , we get by N |= ψ3 that α + ∈ N , whence α + ∈ K  . Hereby K  must be a limit ordinal. Now we have VK  = {Vα : α < K  }, and for each α < K  we have Vα ⊆ N by the facts that Vα ∈ N and N is transitive. Therefore VK  ⊆ M. On the other hand, by N |= ψ5 , N ⊆ VK  . Thus N = VK  . This, combined with the previous results, yields |VK  | = |K  |, which in turn with the fact that ω < K  shows that K  is uncountable. Therefore ω + K  = K  , whence  K  = |Vω+K  | = |VK  | = |K  |. Thus K  is a cardinal and K  =  K  . We have shown that M is isomorphic to N = (Vκ , ∈, K  , V  , f  ) for some κ = κ .

Chapter 8 Exercise 8.3 Because (total) independence and (total) dependence are defined in terms of determination, we must first find the minimal (in respect of ⊆) sets W ⊆ {x2 , x3 } that determine {x1 }. First consider {x2 }. If {x2 } ≥ {x1 } was to hold, it would mean the existence of a function f that maps 1 → 0 (as required by agent s1 of the team), 0 → 0 (as required by s2 ), and 1 → 1 (as required by s3 ). But such a function cannot exist as the requirements by agents s1 and s3 are contradictory. (By the way, this definition of determination in terms of functions is equivalent to the one given in the course material, as one can prove.) Thus {x2 } ≥ {x1 }. Now consider {x3 }. Because there is no function that maps 0 → 0, 1 → 0, and 1 → 1, we can conclude that {x3 } ≥ {x1 }. Now consider {x2 , x3 }. To establish {x2 , x3 } ≥ {x1 }, we need to find a function that maps (1, 0) → 0, (0, 1) → 0, and (1, 1) → 1. But this is possible, so we have {x2 , x3 } ≥ {x1 }. By these three observations we can now say that {x2 , x3 } is the only minimal subset of {x2 , x3 } that determines {x1 }. In fact, it is the only such subset, not only the only minimal one. Now we can say that {x1 } is dependent on {x2 , x3 }, {x2 }, {x3 }, and ∅, i.e. all subsets of {x2 , x3 }. This is so because each of these is included in some minimal set that determines {x1 }, namely {x2 , x3 }.

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Solutions to selected exercises, by Ville Nurmi

We can see that {x1 } is also totally dependent on {x2 , x3 }, {x2 }, {x3 }, and ∅, because each of them is included in all of the minimal determining sets. Well, there was only one minimal determining set in the first place, namely {x2 , x3 }. Note that{x1 } is independent of ∅ and nothing else because only for ∅ can we find a minimal determining set that it avoids. All other W ⊆ {x2 , x3 } intersect all the minimal sets that determine {x1 }. Note that {x1 } is also totally independent of ∅ and nothing else because of the same reason why it is independent of only that set, and the fact that {x2 , x3 } is the only minimal set that determines {x1 }.

Exercise 8.4 By concentrated checking of all possible choices – as in Exercise 8.3 – we can find out that the only minimal sets W ⊆ {x2 , x3 , x4 , x5 } that determine {x1 } are W1 = {x2 , x3 } and W2 = {x3 , x4 , x5 }. All the other sets either do not determine {x1 } or they include W1 or W2 . Now to the actual questions. Note that {x1 } is dependent on every set that is included in some minimal set that determines {x1 }; i.e. {x1 } is dependent on every W ⊆ W1 and every W ⊆ W2 , i.e. {x2 , x3 }, {x2 }, {x3 }, {x3 , x4 , x5 }, {x3 , x4 }, {x3 , x5 }, {x4 , x5 }, {x4 }, {x5 }, and ∅. Note that {x1 } is totally dependent on every set W that is included in all the sets that determine {x1 }. In fact it suffices that this W is included in all minimal determining sets, because every determining set contains a minimal determining set, which then contains W . Thus, {x1 } is totally dependent on every W ⊆ W1 ∩ W2 , that is, {x3 }, and ∅. Note that {x1 } is independent of all the sets that do not intersect some set that determines {x1 }. If a set does not intersect some determining set, then it does not intersect some minimal determining set either. Thus {x1 } is independent of every W ⊆ {x2 , x3 , x4 , x5 } \ W1 and every W ⊆ {x2 , x3 , x4 , x5 } \ W2 , i.e. {x4 , x5 }, {x4 }, {x5 }, {x2 }, and ∅. Note that {x1 } is totally independent of every set that does not intersect  any minimal set thatdetermines {x1 }; i.e. {x1 } is totally independent of every W ⊆ {x2 , x3 , x4 , x5 } \ W1 ∩ {x2 , x3 , x4 , x5 } \ W2 , i.e. ∅ alone.

Exercise 8.5 If, for example, a = 0 then {x3 , x4 } determines {x1 }. This is verified by the mapping f : (3, 6) → 15, (1, 1) → 1, (2, 0) → 6, (2, 2) → 4, (0, 5) → 0. If a = 2 then {x3 , x4 } does not determine {x1 } because s3 and s4 are identical in {x3 , x4 } but different in {x1 }. Or, in other words, there is no function that maps (3, 6) → 15, (1, 1) → 1, (2, 2) → 6, (2, 2) → 4, and (0, 5) → 0. If, for example, a = 0 then {x2 , x4 } determines {x1 }. This is verified by the mapping g : (15, 6) → 15, (1, 1) → 1, (0, 0) → 6, (3, 2) → 4, (0, 5) → 0. If a = 5 then {x2 , x4 } does not determine {x1 } because s3 and s5 are identical in {x2 , x4 } but different in {x1 }. Or, in other words, there is no function that maps (15, 6) → 15, (1, 1) → 1, (0, 5) → 6, (3, 2) → 4, and (0, 5) → 0.

Exercise 8.6 In team (a) in Table A.15, {x1 } determines {x4 }, and so does {x3 }. Therefore {x4 } is dependent but not totally dependent on {x1 }.

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215

Table A.15. Teams (a), (b), (c), and (d) for Exercise 8.6 (a)

s1 s2 s3

(b)

x1

x2

x3

x4

0 0 1

0 1 1

0 0 1

0 0 1

s1 s2 s3

x1

x2

x3

x4

0 0 1

0 1 1

0 0 0

0 0 1

(c)

s1 s2 s3

(d)

x1

x2

x3

x4

0 0 1

0 1 1

0 1 0

0 1 1

s1 s2 s3

x1

x2

x3

x4

0 0 1

0 1 1

0 0 0

0 1 1

In team (b), {x1 } is the only minimal W ⊆ {x1 , x2 , x3 } that determines {x4 }. Therefore {x4 } is totally dependent on {x1 }. In team (c), {x2 } is a minimal set that determines {x4 }, and so is {x1 , x3 }. Because {x1 } intersects only the latter one of these two, {x4 } is independent but not totally independent of {x1 }. In team (d), {x2 } is the only minimal W ⊆ {x1 , x2 , x3 } that determines {x4 }, and {x1 } does not intersect in it. Therefore {x4 } is totally independent of {x1 }. Note that there exist also other correct answers, not only the ones shown here.

Exercise 8.7 We want to prove for all TL-formulas φ where ∼ does not occur, all models M and all teams X with Fr(φ) ⊆ dom(X ) that there is a D-formula φ ∗ such that M |= X φ ⇐⇒ M |= X φ ∗ .

(A.39)

Note that on the left side of Eq. (A.39) we use TL-semantics and on the right side we use D-semantics. We prove the claim by induction on φ by using the following translation from ∼-free TL-formulas to D-formulas: (t = t  )∗ → t = t  ,  ∗

(φ ⊗ ψ)∗ → φ ∗ ∨ ψ ∗ , 

(¬t = t ) → ¬t = t ,

(φ ∧ ψ)∗ → φ ∗ ∧ ψ ∗ ,

(Rt1 . . . tn )∗ → Rt1 . . . tn ,

(∃xn φ)∗ → ∃xn φ ∗ ,

∗

(¬Rt1 . . . tn ) → ¬Rt1 . . . tn ,

(!xn φ)∗ → ∀xn φ ∗ .

(=(t1 , . . . , tn ))∗ → =(t1 , . . . , tn ), The claim in Eq. (A.39) can now be proved by following definitions because the TL-semantics of φ is exactly the same as the D-semantics of φ ∗ .

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Solutions to selected exercises, by Ville Nurmi

Exercise 8.8 Recall the definitions of the four dependence values:  is =();

⊥ is ∼ =();

0 is ¬ =();

1 is ∼ ¬ =().

To prove items (i) and (ii) of Example 8.7, first note that M |= X =() holds for all teams X . This gives 0 ⇒  and 1 ⇒ . We also get that M |= X ∼ =() holds for no teams X . This gives ⊥ ⇒ 0 and ⊥ ⇒ 1. Then we prove items (iii) and (iv). We see immediately that 1 = ∼ 0 and ⊥ = ∼ . It is also the case that M |= X ∼ ∼ φ iff M |= X φ for any TL-formula φ. Therefore we get 0 ≡ ∼ ∼ 0 = ∼ 1, and  ≡ ∼ ∼  = ∼ ⊥. Finally, for item (v), we observe that 0 = ¬ by definition.

Exercise 8.9 Let M be a model and let X be a team for M. Because M |= X  holds (for any team), we get M |= X φ ∧  ⇐⇒ M |= X φ and M |= X  ⇐⇒ M |= X φ. Recall that φ ∨ ψ that is shorthand for ∼(∼ φ ∧ ∼ ψ) and that φ ⊕ ψ is shorthand for ∼(∼ φ ⊗ ∼ ψ). We have the following chains of equivalences: M |= X ∼(∼ φ ∧ ∼ ) ⇐⇒ M |= X ∼ φ ∧ ∼  ⇐⇒ M |= X ∼ φ or M |= X ∼  ⇐⇒ M |= X φ or M |= X  ⇐⇒ M |= X . Note that when a team X is given, and Y ∪ Z = X holds for some sets Y and Z , then dom(Y ) = dom(Z ) necessarily holds. Thus in these cases we can omit this condition. Now, M |= X  ⊗  ⇐⇒ M |=Y  and M |= Z  for some Y ∪ Z = X ⇐⇒ M |= X . The last equivalence above holds because both sides are always true. Further, M |= X ∼(∼ φ ⊗ ∼ ) ⇐⇒ M |= X ∼ φ ⊗ ∼  ⇐⇒ for all Y ∪ Z = X : M |=Y ∼ φ or M |= Z ∼  ⇐⇒ for all Y ∪ Z = X : M |=Y φ or M |= Z  ⇐⇒ M |= X . The last equivalence above holds because both sides are always true.

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217

Exercise 8.10 Let M be a model and X a team for M. Because M |= X ⊥ does not hold (for any team), we get M |= X φ ∧ ⊥ ⇐⇒ M |= X φ and M |= X ⊥ ⇐⇒ M |= X ⊥. Recall that φ ∨ ψ is shorthand for ∼(∼ φ ∧ ∼ ψ) and that φ ⊕ ψ is shorthand for ∼(∼ φ ⊗ ∼ ψ). We have the following chains of equivalences: M |= X ∼(∼ φ ∧ ∼ ⊥) ⇐⇒

M |= X ∼ φ ∧ ∼ ⊥

⇐⇒

M |= X ∼ φ or M |= X ∼ ⊥

⇐⇒

M |= X φ or M |= X ⊥

⇐⇒

M |= X φ;

and M |= X φ ⊗ ⊥ ⇐⇒

M |=Y φ and M |= Z ⊥ for some Y ∪ Z = X

⇐⇒

M |= X ⊥.

The last equivalence above holds because both sides are always false. Finally, M |= X ∼(∼ ⊥ ⊗ ∼ ⊥) ⇐⇒ M |= X ∼ ⊥ ⊗ ∼ ⊥ ⇐⇒ for all Y ∪ Z = X : M |=Y ∼ ⊥ or M |= Z ∼ ⊥ ⇐⇒ for all Y ∪ Z = X : M |=Y ⊥ or M |= Z ⊥ ⇐⇒ M |= X ⊥. The last equivalence above holds because both sides are always false.

Exercise 8.11 Let M be a model and let X be a team for M. Assume first that φ is a D-formula (that is, φ is a TL-formula, and ∼ does not occur in φ). Then we have the old result that M |=∅ φ. Also recall that M |= X 0 iff X = ∅. With these observations we get the following chain of equivalences: M |= X φ ∧ 0 ⇐⇒ M |= X φ and M |= X 0

(A.40)

⇐⇒ X = ∅

(A.41)

⇐⇒ M |= X 0.

218

Solutions to selected exercises, by Ville Nurmi

The implication Eq. (A.41) ⇒ Eq. (A.40) is based on φ being a D-formula. Also, M |= X ∼(∼ φ ∧ ∼ 0) ⇐⇒ M |= X ∼ φ ∧ ∼ 0 ⇐⇒ M |= X ∼ φ or M |= X ∼ 0 ⇐⇒ M |= X φ or M |= X 0 ⇐⇒ M |= X φ or X = ∅

(A.42)

⇐⇒ M |= X φ.

(A.43)

The implication Eq. (A.42) ⇒ Eq. (A.43) is based on φ being a D-formula: Now we let φ be any TL-formula: M |= X φ ⊗ 0 ⇐⇒ M |=Y φ and M |= Z 0 for some Y ∪ Z = X ⇐⇒ M |=Y φ and Z = ∅ for some Y ∪ Z = X ⇐⇒ M |= X φ;

M |= X ∼(∼ 0 ⊗ ∼ 0) ⇐⇒ for all Y ∪ Z = X : M |=Y ∼ 0 or M |= Z ∼ 0 ⇐⇒ for all Y ∪ Z = X : M |=Y 0 or M |= Z 0 ⇐⇒ for all Y ∪ Z = X : Y = ∅ or Z = ∅ ⇐⇒ X = ∅ ⇐⇒ M |= X 0.

Exercise 8.13 Let X = {∅} and let M be a model with at least two elements. Choose φ to be the TL-formula 1 ⊗ 1. Then M |= X !x0 φ. To see why, observe that M |= X !x0 φ iff M |=Y φ, where Y = X (M/x0 ) has at least two elements (as many as M has). We can split Y = Y1 ∪ Y2 so that both Y1 and Y2 are non-empty, whence M |=Yi 1 for both i = 1, 2. Thus M |=Y φ. On the other hand, M |= X ∀x0 φ does not hold. To see why, observe that M |= X ∀x0 φ iff M |= Z φ, where Z = X (F/x0 ) has only one element (as many as X has). Thus any split Z = Z 1 ∪ Z 2 is bound to have Z i = ∅ for one i = 1, 2. For this i we have M |= Z i 1, so we get M |= Z φ.

Exercise 8.15 Let φ be a TL-formula, let M be a model, and let X be a team for M. We want to express that for all subsets Y ⊆ X there is a subset Z ⊆ Y for which φ holds. Let ψ be

Solutions to selected exercises, by Ville Nurmi the TL-formula (φ ⊗ ) ⊕ ⊥. Then we have the following: M |= X ψ ⇐⇒ for all Y ∪ Z = X : M |=Y φ ⊗  or M |= Z ⊥ ⇐⇒ for all Y ⊆ X : M |=Y φ ⊗  ⇐⇒ for all Y ⊆ X there is Z ∪ V = Y : M |= Z φ and M |=V  ⇐⇒ for all Y ⊆ X there is Z ⊆ Y : M |= Z φ.

219

References

[1]

[2] [3] [4]

[5] [6] [7] [8] [9] [10]

[11] [12] [13] [14]

W. W. Armstrong. Dependency structures of data base relationships. In Information Processing 74. Proc. IFIP Congress, Stockholm (Amsterdam: North-Holland, 1974), pp. 580–583. A. Blass and Y. Gurevich. Henkin quantifiers and complete problems. Ann. Pure Appl. Logic, 32:3 (1986), 1–16. J. P. Burgess. A remark on Henkin sentences and their contraries. Notre Dame J. Formal Logic, 44:3 (2003), 185–188. X. Caicedo and M. Krynicki. Quantifiers for reasoning with imperfect information and 11 -logic. In W. A. Carnielli and I. M. L. D’Ottaviano, eds., Advances in Contemporary Logic and Computer Science (Providence, RI: American Mathematical Society, 1999), pp. 17–31. 1999. P. Cameron and W. Hodges. Some combinatorics of imperfect information. J. Symbolic Logic, 66:2 (2001), 673–684. W. Craig. Linear reasoning. A new form of the Herbrand-Gentzen theorem. J. Symbolic Logic, 22 (1957), 250–268. H. B. Enderton. Finite partially-ordered quantifiers. Z. Math. Logik Grundlagen Math., 16, (1970), 393–397. H. B. Enderton. A Mathematical Introduction to Logic, 2nd edn. (Burlington, MA: Harcourt/Academic Press, 2001). T. Frayne, A. C. Morel, and D. S. Scott. Reduced direct products. Fund. Math., 51, (1962/3), 195–228. D. Gale and F. M. Stewart. Infinite games with perfect information. In H. W. Kuhn and A. W. Tucker, eds., Contributions to the Theory of Games, vol. 2, (Princeton: Princeton University Press, 1953), pp. 245–266. J.-Y. Girard. Linear logic. Theor. Comp. Sci., 50:1 (1987), 101 pp. K. G¨odel. Collected Works. Vol. I. (New York: Oxford University Press, 1986), Publications 1929–1936. Edited and with a preface by Solomon Feferman. D. Harel. Characterizing second-order logic with first-order quantifiers, Z. Math. Logik Grundlag. Math., 25:5 (1979), 419–422. L. Henkin. Some remarks on infinitely long formulas. In Infinitistic Methods. Proc. Symposium Foundations of Math., Warsaw, (London: Pergamon, 1961), pp. 167–183.

220

References

221

[15] L. Henkin. The completeness of the first-order functional calculus. J. Symbolic Logic, 14 (1949), 159–166. [16] J. Hintikka and G. Sandu. Informational independence as a semantical phenomenon. In J. E. Fenstad, I. T. Frolov, and R. Hilpinen, eds., Logic, Methodology and Philosophy of Science, VIII. Stud. Logic Found. Math., vol. 126 (Amsterdam: North-Holland, 1989), pp. 571–589. [17] J. Hintikka and G. Sandu. Game-theoretic semantics. In J. F. A. K. van Benthem and G. B. A. ter Meulen, eds., Handbook of Logic and Language, vol. 1 (Boston, MA: MIT Press, 1997), pp. 361–410. [18] J. Hintikka. Form and content in quantification theory. Acta Philos. Fenn., 8, (1955), 7–55. [19] J. Hintikka. The Principles of Mathematics Revisited (Cambridge: Cambridge University Press, 1996). [20] J. Hintikka. Hyperclassical logic (a.k.a. IF logic) and its implications for logical theory. Bull. Symbolic Logic, 8:3 (2002), 404–423. [21] W. Hodges. Compositional semantics for a language of imperfect information. Logic J. IGPL, 5:4 (1997), 539–563. [22] W. Hodges. Some strange quantifiers. In J. Mycielski, G. Rozenberg, and A. Salomaa, eds., Structures in Logic and Computer Science. Lecture Notes in Computer Sci., vol. 1261 (Berlin: Springer, 1997), pp. 51–65. [23] T. Janssen and F. Dechesne. Signalling: a tricky business. In J. van Benthem, G. Heinzmann, M. Rebuschi, and H. Visser, eds., The Age of Alternative Logics: Assessing Philosophy of Logic and Mathematics Today (Berlin: Springer, 2006). [24] T. M. V. Janssen. Independent choices and the interpretation of IF logic. J. Logic Lang. Inform., 11:3 (2002), 367–387. [25] T. Jech. Set Theory, 2nd edn., (Berlin: Springer-Verlag, 1997). [26] S. Krajewski. Mutually inconsistent satisfaction classes. Bull. Acad. Polon. Sci. S´er. Sci. Math. Astron. Phys., 22 (1974), 983–987. [27] G. Kreisel and J.-L. Krivine. Elements of Mathematical Logic. Model Theory. (Amsterdam: North-Holland Publishing Co., 1967). [28] S. Kripke. An outline of a theory of truth. J. Phil., 72 (1975), 690–715. [29] M. Krynicki and A. H. Lachlan. On the semantics of the Henkin quantifier, J. Symbolic Logic, 44:2 1999), 184–200. [30] A. L´evy. A hierarchy of formulas in set theory. Mem. Am. Math. Soc., 57 (1965), 76 pp. [31] E. G. K. Lopez-Escobar. A non-interpolation theorem. Bull. Acad. Polon. Sci. S´er. Sci. Math. Astronom. Phys., 17:109–112, 1969. [32] M. Magidor. On the role of supercompact and extendible cardinals in logic. Israel J. Math., 10 (1971), 147–157. [33] Y. N. Moschovakis. Elementary Induction on Abstract Structures. Studies in Logic and the Foundations of Mathematics, vol. 77 (Amsterdam: North-Holland, 1974). [34] A. Robinson. A result on consistency and its application to the theory of definition. Nederl. Akad. Wetensch. Proc. Ser. A. Math., 18 (1956), 47–58. [35] G. Sandu and T. Hyttinen. IF logic and the foundations of mathematics. Synthese, 126: 1–2 (2001), 37–47. [36] T. Skolem. Logisch-kombinatorische Untersuchungen u¨ ber die Erf¨ullbarkeit oder Beweisbarkeit mathematischer S¨atze nebst einem Theoreme u¨ ber dichte

222

[37] [38] [39] [40] [41] [42] [43] [44]

References

Mengen. Videnskapsselskapets skrifter I, Matematisknaturvidenskabelig Klasse, Videnskappsselskapet i Kristiania, 4 (1920), 1–36. T. Skolem. Selected Works in Logic (Oslo: Scandinavian University Press, 1970). A. Tarski. The semantic conception of truth and the foundations of semantics. Phil. Phenomenol. Res., 4 (1944), 341–376. A. Tarski and R. L. Vaught. Arithmetical extensions of relational systems. Compositio Math, 13 (1958), 81–102. J. V¨aa¨ n¨anen. Second-order logic and foundations of mathematics. Bull. Symbolic Logic, 7:4 (2001), 504–520. J. V¨aa¨ n¨anen. On the semantics of informational independence. Logic J. IGPL, 10:3 (2002), 339–352. J. V¨aa¨ n¨anen. A remark on nondeterminacy in IF logic. Acta Philosophica Fennica, 78 (2006), 71–77. J. von Neumann and O. Morgenstern. Theory of Games and Economic Behavior (Princeton: Princeton University Press, 1944). W. J. Walkoe, Jr. Finite partially-ordered quantification. J. Symbolic Logic, 35 (1970), 535–555.

Index

Abelard (player I) 63 abelian group 53 absoluteness 140 agent 16 α ∗ 63 α 63 anti-symmetric relation 56 arithmetical relation 107 arity 5 assignment 6 atomic formula 6, 18 automorphism 52 avoiding a model 134 Axiom of Extensionality 60, 162 backslashed quantifier 44 Beth Definability Theorem 100 Beth numbers 137 binary tree 56 bounded quantifiers 138 chain 56 Chinese Remainder Theorem 104 clique 57 codomain 16 cofinality 56 Compactness Theorem 91 completeness 53 compositionality 32 connected graph 56 consistency 118 Consistency Problem 134 consistency property 111 constant symbol 5 constant term 6

Continuum Hypothesis 139 Craig Interpolation Theorem interpolation 98 cumulative hierarchy 61 cycle 57 D 16 D-equivalence 121 D-semiequivalence 122 decision problem 134 definability 100 degree 52 dependence back-and-forth set 132 dependence friendly logic 45 dependence logic 3, 10, 16 dependence values 154 determined game 64, 92 DF see dependence friendly logic disconnected graph 56 domain 5, 16 duplicate 20 Eloise (player II) 63 equicardinality 52 equivalence relation 57 expansion 6 extended IF logic 144 false sentence 21 feature 16 first order logic 19 Flatness Test 39 flattening 42 free variable 7, 19 function symbol 5 fundamental predicate 20

223

224

Gale–Stewart Theorem 64 game tree 64 generated team 12 G¨odel’s Incompleteness Theorem 101 G¨odel-number 101 group 53 guessing 53 half-round 70 Hamiltonian 57 Henkin quantifiers 99 Hintikka set 113 hyperarithmetical relation 107 IF see independence friendly logic ill-founded relation 55 imperfect information 63 inaccessible cardinal 139 inconsistency 134 Inconsistency Problem 134 independence friendly logic 46 independent sentence 101 information set 63 inseparable sets 119 isolated vertex 52 isomorphism 6 #L 5 L¨owenheim–Skolem Theorem 91 Law of Excluded Middle 3, 4, 100 Levy Hierarchy 137 Liar Paradox 100, 108 measurable cardinal 139 model class 128 model existence game 110, 117 modified assignment 7 Mostowski’s Collapsing Lemma 140 negation normal form 7 non-determined game 92 non-valid 134 Non-validity Problem 134 normal form 98 omitting a type 120 Omitting Types Theorem 120 opponent 63 Padoa Principle 100 partial dependence isomorphism 132

Index

partial order 55 partial play 64 path 57 PC definition 87 Peano’s axioms 102 perfect information 64 = 52 ∼ = 52 ∞ 51 ≤ 52 cml 54 conn 56 even 48 ext 60 nr 53  N 59 N 58 R 59 R,N 60 set 61 wf 55 play 64 prenex normal form 44 principal type 120 principle of duality 9 rank 136 recursively isomorphic sets 162 reduct 6 reflexive relation 55 relation symbol 5 right orderable group 53 rigid structure 53 root 56 satisfaction 7 satisfying a formula 120 second order logic 158 semantic proof 117 sentence 7, 19 separable sets 119 Separation Theorem 91 shriek 151 11 86 signal 46 Skolem Normal Form 94 slashed quantifier 45 standard model 106 strategy 65, 71, 72, 81 use of 67 structure 5

Index

subtree 56 supplement 19

-tuple 5 type 17, 21, 120

t M s 6 team 16 team logic 144, 151 tensor 151 term 6 3-colorable 57 transitive closure 136 transitive relation 56 transitive set 136 tree 56 true sentence 21 truth-definition 100 truth-table method 43

uniform strategy 82 uniform interpolant 98 Uniform Interpolation Property 98 universal-existential formula 98 valid sentence 21 variable 5 veritas symbol 6, 19 vocabulary 5 well-founded relation 55 winning strategy 66, 73, 82 zero-sum 64

225